We have to repair the CFG to be able to derive target sequence: "1 1 25 17 14 16 23 22 21 20 24 1 15 18 19 17 13 3 11 2 10 6 7 9 8 1 12 4 5 3". Actually we need to output the updated production rules after inserting terminal symbols minimal number of insertions.

Observations: The grammar appears heavily design with repeated rules for A2->A58 A3 and A2->..., but the target includes many terminals. Let's see the derivation path desired: Starting symbol A1 -> A2 (rule1). A2 -> A58 A3 (rule 2). So the string is derived from concatenating derivations of A58 and A3.

Goal: produce the target sequence: A sequence of length 30 (from counting) digits: Let me list them:

Indices:

1: 1
2: 1
3: 25
4: 17
5: 14
6: 16
7: 23
8: 22
9: 21
10: 20
11: 24
12: 1
13: 15
14: 18
15: 19
16: 17
17: 13
18: 3
19: 11
20: 2
21: 10
22: 6
23: 7
24: 9
25: 8
26: 1
27: 12
28: 4
29: 5
30: 3

All terminals from 1 to 25 appear at least once. The grammar has many productions where terminals appear as sequences embedded with non-terminals.

But many rules have omitted terminals that need to be added. Example: A58 -> A60 A59, where A60 and A59 have productions with only 1's, but missing other terminals like 25 etc.

The A9-> 25 14 16 A16 15 18 19 17 (rule 427). This matches part of the target: "25 14 16 ... 15 18 19 17". Indeed rule 427: A9 -> 25 14 16 A16 15 18 19 17. Expand A16 to produce a terminal maybe 23?? Actually A16->1 or A16 -> A17 A16; A17 -> 20|21|22|23|24. So we can produce a terminal 23... Actually we need after 16 the target sequence includes 23,22,21,20,24 then 1 then 15... Wait check: after 16 we have "23 22 21 20 24 1". So "A16" must produce "23 22 21 20 24 1"? Let's examine A16 definition: A16 -> 1; A16 -> A17 A16. So A16 can produce sequences of A17 repeated, ending with 1. Each A17 can produce 20,21,22,23,24, each singly. So we can get any sequence of these numbers ending with 1, like 23 22 21 20 24 1 (yes!). So rule 427 yields prefix "25 14 16" plus A16's produced sequence "23 22 21 20 24 1" then continue "15 18 19 17". So that matches part of target: the sequence 25 14 16 23 22 21 20 24 1 15 18 19 17.

Now after that target continues: "13 3 11 2 10". The remaining part is "13 3 11 2 10 6 7 9 8 1 12 4 5 3". So after the above segment we have "13". In grammar somewhere we have A13 -> 13 3 11 2 A14 5 3 (rule 444). That seems promising. Let's see: A13 contains "13 3 11 2 ..." which matches "13 3 11 2". Then A14, then "5 3". Need to match "10 6 7 9 8 1 12 4 5 3". Actually after "13 3 11 2", we have "10 6 7 9 8 1 12 4 5 3". Let's inspect A14: A14 -> 1, A14 -> A15 A14. So A14 can produce 1 repeated some number of times (plus maybe expansions). A15 can be 6,7,8,9,10. So maybe we can generate "10 6 7 9 8 1 12 4". Actually A14 cannot generate 10 directly; it's just A15 and A14 recursion.

Define A14 yields a sequence of A15's ending with 1. Since A15 productions: 6, 7, 8, 9, 10. So A14 -> A15 A14 can produce any sequence of those numbers (including duplicates) ending with 1? Actually recursion: A14 -> A15 A14. Base case A14 -> 1. So A14 can produce one or more A15's then finally a 1, but the order is: A15 then A14 (which may then produce more A15s), so you get something like (A15 ...) ... 1 at end.

Thus A14 can generate any string of numbers from {6,7,8,9,10} ending with 1. Eg to get "10 6 7 9 8 1" is possible by A14 => A15 A14 (choose 10) => A14 => A15 A14 (choose 6) =>... etc eventually base case 1. So that matches.

Now after A14 in rule 444 we have "5 3". So we have "5 3" after the A14 sequence; but target after that is "... 12 4 5 3". So we still need "12 4" before the "5 3". Possibly A14 could produce "10 6 7 9 8 1 12 4"? But A14 cannot produce 12 or 4 because A15 only yields 6-10, not 12 or 4; A14 introduces only 1 at end. To get "12 4", we need further expansions after A14 maybe using A13's rest.

Search for other A13 productions: A13 -> 1 (rule 441); A13 -> A7 A5; A13 -> A8 A13; A13 -> 13 3 11 2 A14 5 3 (rule 444); A13 -> A13 A12 (rule 445). A13 also may produce something else but those are limited.

The rule where we have "A13 -> A13 A12" could produce A13 (some pattern) followed by A12; A12 expansions: A12 -> 1; A12 -> A10 A13; A12 -> A11 A5; A12 -> A12 A12. So A12 can produce "1" or other patterns to get may be "12 4"? Not directly; but A12 is a nonterminal to generate maybe 12 and/or 4.

Check A10: A10 -> 1; A10 -> A10 A8; A10 -> A11 A9; A10 -> A12 A10. So A10 can produce many sequences including A8, A11 etc.

A11: A11 -> 1; A11 -> A10 A7; A11 -> A11 A6; A11 -> A12 A11.

A7 -> productions that include "1", recursion, and "A8 A7" and "A13 A11". So A7 can produce many patterns.

But maybe we can get the remaining "12 4 5 3" using some other production after A13 uses A14. Possibly we need to insert missing terminals into some rule to generate "12" and "4". Both are terminals, missing from some productions, as we see many rules where just "1" or "13" but not "12" or "4". The target includes 12 and 4. However productions that include 12 are A66->12 (rule 1954) and maybe others. Production of A16 also includes 1 and the recursion uses A17 A16, but A17 yields 20-24 only. No 12 there.

Productions containing 4 as a terminal: I see rule 1120 (A33 -> A33 A32) not contain numbers. There is rule 1134: A37 -> 13 11 A14 5 (ends with 5). There's rule 1135: A37 -> 13 11 A14 12 5 3 (includes 12 and 5 and 3). That's promising. A37 may produce "13 11 A14 12 5 3". After "13 3 11 2 ..." we need later "10 6 7 9 8 1 12 4 5 3". This contains "13 11 A14 12 5 3"? Actually we need the 13 again; we have 13 earlier inside A13 rule. There's also A37 which contains 13 11 A14 12 5 3. Yet the target after the segment we had after rule 444: we have "13 ...". Actually we have already produced "13" from A13, but maybe we need another production that yields part of remainder: "13 3 11 2 10 ...". Let's try to identify path.

Potential combination: The target could be derived by A9 (rule 427) yields the first segment up to "... 15 18 19 17". Then maybe A3 -> A38? Wait A3 -> A28? Actually rule 102: A3 -> A4; rule 103: A3 -> A18; rule 104: A3 -> A28; rule 105: A3 -> A38; rule 106: A3 -> A48. So after A9 part (which ends with terminal 17) we go to A3, then maybe choose A3 -> A38 produce remainder. Check A38-> A39..., eventually A47 -> includes pattern "13 2 A14 5 3" etc.

But after A9, we need to produce: "13 3 11 2 10 ...". Actually after "15 18 19 17" target continues "13 3 11 2 10". So maybe we need to generate "13 3 11 2 10 6 7 9 8 1 12 4 5 3". Could be via A38-> A39... or A28 etc.

Let's analyze A38 productions: they all produce A39 (which leads eventually to A47). A39 leads to something: A39 -> A40 A39, etc. Finally there are productions with terminals: A43 -> 25 14 A16 (and also long sequence includes 25 14 ...). A47 includes many productions: "13 2 A14 5 3", "13 11 A14 5", "13 A14 5". Not "13 3 11 2". However note A9 gave "25 14 16 ..." using A16. But A47 provides "13 2 A14 5 3". Yet we need "13 3 11 2". That's similar but swapped order of 3 and 11.

There is A37 rule 444: "13 3 11 2 A14 5 3" exactly the pattern we need after some prefix maybe.

In A13 rule 444? Wait rule 444: A13 -> 13 3 11 2 A14 5 3. Yes that directly gives "13 3 11 2 A14 5 3". So after we produce "15 18 19 17", perhaps we take A3 (or A5?). Actually A3 -> A4? A3 -> A28? In our derivation A2 -> A58 A3. A58 gave "A60 A59"? Actually A58 -> A60 A59; A60 produces maybe some pattern, but we mainly aren't using that. Actually we need A58's contribution to give the segment before A9 maybe? Let's examine A58's productions: rule 1819: A58 -> A60 A59; rule 1820: A58 -> A63 A61; rule 1821: A58 -> A66 A64.

Terminals defined: A60 has only "1" productions (maybe many duplicates), A59 also only "1". So A58 can produce "1 1" (if choose first rule). Or we could use other rules: A58 -> A63 A61: A63 includes terminal at rule 1907: A63 -> 15, and rule 1908: A63 -> 25. A61 includes many "1". So maybe A58 can produce "15 1". Or A58 -> A66 A64: A66 includes 12 or 13. A64 includes many "1". So can produce e.g., "12 1". So overall, A58 can produce some prefixed terminal segment: we need "1 1" at start of target. The target begins with "1 1". So we can get that from A58 -> A60 A59, both produce 1 each. So that's good.

Thus full derivation:

A1 -> A2.
A2 -> A58 A3.

A58: choose rule_1819: A58 -> A60 A59.
 A60 produces 1 (any of its rules). A59 produces 1.

Then A3: we need to produce the remaining 28 terminals: starting from third terminal "25 17 14 16 23 22 21 20 24 1 15 18 19 17 13 3 11 2 10 6 7 9 8 1 12 4 5 3". Let's see if we can produce this using A3 -> A9? No; A4 etc. Let's check A3 -> A4 (rule 102). A4 has tons of productions: many produce A5, many produce A12, many produce A13. So we can embed sequences via A4 expansions.

Option: Let A4 -> A5 to produce something including 25? Actually A5 -> 1 (rule 407) and also recursion combos.

But there is rule 427: A9 -> 25 14 16 A16 15 18 19 17. That's a perfect segment of the target from 25 up to 17 (the first 17 after 19). Actually target after first two 1's is: 25 17 14 16 ... Wait target sequence list is "25 17 14 16 23 ..."? Actually I need to double-check target (given): "1 1 25 17 14 16 23 22 21 20 24 1 15 18 19 17 13 3 11 2 10 6 7 9 8 1 12 4 5 3". So after two 1's we have 25 17 14 16 ... So there is "17" immediate after 25, before 14. But rule 427 starts with "25 14 16". Lacks the 17 after 25. However there is A43 -> "25 14 A16". A16 could produce "17"? Actually A16 is from earlier (A16 is used in A58?). Wait A16's expansions are: 1 or A17 A16, where A17 yields 20-24. So A16 cannot produce 17; earlier we needed "17" in that location as terminal (target has 17 after 25). However we have other productions: A43 -> 25 14 A16 (but that yields "25 14" then A16, which cannot produce "17" because A16 can produce numbers 20-24 etc plus 1. So maybe we need to produce that 17 using some other rule.

Let's reconsider the target sequence part: 25 17 14 16. Actually we see "25 17 14 16"? That seems unusual. Typically 17 is the terminal for nonterminal A?? Many rules include 17 as a terminal: e.g., A62 -> 17 (rule_1900). Also many A9 rules include numbers "25 14 16". A43 includes 25 14 A16. But not "25 17". Perhaps we could produce "25" then "17" separately, e.g., "25" from some production and "17" from somewhere else.

One candidate: A43 -> 25 14 A16 occurs after we need "25 17 14 16". Possibly we could produce "25" from A5? Not sure.

Another possibility: Use A9 -> "25 14 16 A16 15 18 19 17" but then we must incorporate the 17 after 25. But we can get 17 before by preceding a 17 using some other nonterminal between 25 and 14. For example "25 17 14 16 A16 ..." could be produced by perhaps A43 -> "25 14 A16" preceded by something that generates "17". Or there is a rule "A9 -> 25 A16 19" (rule 428) not involve 14 or 16.

Let's step through target in order and try mapping to productions.

Given the constraints: We can only insert terminal symbols into existing production rules, not reorder or delete. So we need to see where missing terminals need to be inserted.

Goal: represent target string as concatenation of expansions from the start symbol A1.

Given numerous duplicate productions, we can choose particular path to realize target with minimal insertions.

We need to see which parts of target currently match existing sequences exactly; for missing bits, we can insert them.

First two 1's: A58->A60 A59 gives "1 1" (both are 1). Already correct; no insert needed.

Now after that we need "25 17 14 16 ...". Let's find a rule that can produce "25 14 16". That's rule 427: A9 -> 25 14 16 A16 15 18 19 17. This yields "25 14 16", but missing the "17" after 25 and before 14. So we need to insert "17" somewhere. Could insert "17" after the 25 in the rule 427, making it "25 17 14 16 ...". However we cannot modify ordering, but we can insert terminal symbols anywhere in the RHS. So we could modify rule 427 to be "A9 -> 25 17 14 16 A16 15 18 19 17". That adds a terminal "17" after 25. That uses insertion.

Now check after "14 16" we need "23 22 21 20 24 1". That's covered by A16? A16 currently can produce sequences of A17 (20-24) ending with 1, but the order is A17 A16 recursively; the sequence is built as left recursion? Let's see: A16 -> A17 A16. This produces sequence of A17 repeated, then eventually 1 (base). So the order is: first A17, then recursion's A16. So expansions produce e.g., with A17->23 then then A16-> A17 A16... eventually base 1, yields sequence "23 ... ...". So we can generate the "23 22 21 20 24 1" as A17 choices: 23, 22, 21, 20, 24, then base 1. That matches order.

Now after that we need "15 18 19 17". That's after A16's production we have "15 18 19 17" already in rule 427. Indeed rule 427 ends with "15 18 19 17". So that's fine.

Thus by inserting 17 after 25, rule 427 yields substring: "25 17 14 16 23 22 21 20 24 1 15 18 19 17".

Next, after that we need "13 3 11 2 10 6 7 9 8 1 12 4 5 3". At this point we may use A13 -> 13 3 11 2 A14 5 3 (rule 444). This yields "13 3 11 2" a prefix and "5 3" suffix, with A14 inserted in middle.

We need to generate "10 6 7 9 8 1 12 4". Currently A14 can generate a sequence of numbers from {6,7,8,9,10} ending with 1, but cannot generate "12" or "4". So we need to add productions that embed 12,4.

Let's review other productions that could provide "12 4". There is rule 1135: A37 -> 13 11 A14 12 5 3. This yields 13 11 (but note missing 3 after 13). Actually A37 -> 13 11 A14 12 5 3 ... Actually rule 1135 is A37 -> 13 11 A14 12 5 3. That includes 12, but also after A14 we have "12 5 3". However we need "12 4 5 3". So need 4 before 5.

There is rule 795: A9 -> 25 14 16 A16 15 18 19 17 (already used). Another rule 780: A23 -> 25 14 A16 (maybe not). There is rule 780: A23 -> 25 14 A16 (we saw). But not relevant.

Check other rules containing terminal 12 and 4: rule 795 (maybe not). Actually rule 795 is A9-> 25 14 16 A16 15 18 19 17 (has 12?). No.

Rule 795 is a rule which we saw earlier for A9 (the same as 427?). Wait mislabeled: rule 427 is A9 -> 25 14 16 A16 15 18 19 17; But rule 427 is a part. Actually I need to check rule numbers there is duplication. I think 427 corresponds to A9->..., while 795 for other maybe but I need to verify: At top, after rule 423 there is rule 427 etc. Indeed rule 427: "A9 -> 25 14 16 A16 15 18 19 17". So rule 795 is later but we haven't read. Actually after rule 794 leaps to higher numbers. You see around rule 795: A9 -> 25 14 16 ... Then rule 795 indeed seems later maybe a duplicate? Wait we saw at rule 795: "A9 -> 25 14 16 A16 15 18 19 17", but that is exactly same as 427. Might be duplicates.

Nevertheless there is rule 795 (maybe duplicate). Not important.

Terminal 4 appears in many rules: For A57 -> 13 11 2 A14 12 4 3 (rule 1815). So A57 can produce 13 11 2 A14 12 4 3. This includes "13 11 2", then A14, then "12 4 3". That's interesting: but we need "13 3 11 2". The ordering is 13 11 2, missing 3 after 13 and before 11. So we might need to adapt something.

Also A57 appears after A57 may be used through some chain. But note we already have A13 -> 13 3 11 2 A14 5 3: that's close but missing 12 4 before the final 5 3. So we want to produce the remainder "10 6 7 9 8 1 12 4" before the final "5 3". The A14 part can generate "10 6 7 9 8 1". Then we need "12 4". So maybe we can modify A14 production to insert a "12 4" at appropriate point: Since A14 can only produce sequences of A15 and 1. However we could insert "12 4" within A14's production via insertion of terminals after the recursion or before base? Let's explore.

A14 rules currently:

- rule_446: A14 -> 1
- rule_447: A14 -> A15 A14

Thus A14 can derive sequence of A15's followed by a base 1. This means the sequence ends with 1. If we want to have "12 4" after the A14 part but before 5 3, we need them inserted after the A14's final 1, before 5. However A14's final 1 is part of its own output; we can't extend after that because A14 is itself a subcomponent of A13's RHS. A13's RHS: "13 3 11 2 A14 5 3". So after A14 we have "5 3". So we can modify rule 444 to insert "12 4" after A14: A13 -> 13 3 11 2 A14 12 4 5 3. This adds terminals "12 4". That is allowed: inserting terminal symbols anywhere in the RHS. So we can augment A13's production (rule 444) via insertion of 12 and 4 after A14 and before 5,3.

But we need to be careful: The rule currently is: A13 -> 13 3 11 2 A14 5 3. Insert "12 4" after A14 yields: A13 -> 13 3 11 2 A14 12 4 5 3. That adds two terminals.

Now does A14 produce the needed "10 6 7 9 8 1"? A14 currently can generate a sequence of A15s ending with 1. A15 can produce 6,7,8,9,10. However the order currently: if we want "10 6 7 9 8 1", we need to properly choose A15s. Yes we can generate that.

Thus with the insertion of "12" and "4" into rule 444 we can produce the final "12 4" portion. So total number of inserted terminals: we need to add "17" after 25 in rule 427 (1 terminal), and "12" and "4" after A14 in rule 444 (2 terminals). Is there any other mismatch? Let's check the target also includes terminal "3" after 5 at the end: the final "3" is already in rule 444 (the final 3). So OK.

Now check earlier we inserted "17" after 25. However we must confirm the resulting string sequence order: Starting path:

A1 -> A2

A2 -> A58 A3

A58 -> A60 A59 (both produce "1")

Thus we have initial "1 1".

Now A3 will produce the rest: Need to select a derivation from A3 that uses A9 and A13.

A3 options: -> A4, A18, A28, A38, A48. Choose A3 -> A4 (rule_102). Then A4 produce something that includes A9 and A13. A4 has many productions: many A5, many A12, many A13 productions. So we can envisage A4 -> A5 (?) Actually A5 yields only 1 or other symbols (nonterminals). But we need to produce "25 17 14 16 ..." So we need A4 to go to a derivation that yields A9 (which contains the segment). There is production A4 -> A5 (most of them). However A5 doesn't produce 25 etc; but A5 can produce via recursion "A5 A12" etc. That could yield terminal sequences 1 and maybe produce A12 expansions etc. That's complicated.

But there is also an A4 -> A12 production. A12 can produce various combos, maybe includes A9? Not directly. Also A4 -> A13 productions (there are many of them). But A13 is the later segment after the preceding part. So we could structure A4 -> A9 then later something else. Actually check A4 productions: rule_107: A4 -> A5; rule_207: A4 -> A12; rule_307: A4 -> A13. So A4's RHS can be any of those.

Actually we need to embed both A9 and A13 in order. Since they are each can produce part of string. We can generate them sequentially via iterative A4 expansions? Not directly because A4's RHS is a single non-terminal; cannot produce concatenation of two non-terminals. However there are also recursive productions for A4? No, A4 is not defined recursively; each A4 rule is just A5, A12, or A13. So to produce a string consisting of concatenation of A9 and A13, we may need to use other non-terminals that embed them.

Potential chain: Use A5 expansions that eventually produce A13 after A9. Let's examine A5 definitions: rule_407: A5 -> 1
rule_408: A5 -> A5 A12
rule_409: A5 -> A6 A5
rule_410: A5 -> A9 A13

Thus A5 can produce A9 A13 as in rule_410. This is perfect! A5 directly contains the concatenation of A9 and A13.

Thus we can produce the segment after the first two 1's as follows: A3 -> A4 (or maybe we don't need A4 at all?). Actually A3 -> A4 could produce A4 -> A5 (via rule_107). Then A5 -> A9 A13 via rule_410. Perfect.

Thus overall derivation:

A1 -> A2

A2 -> A58 A3

A58 -> A60 A59 (both -> 1) => "1 1". Good.

Now A3 -> A4 (rule 102).

A4 -> A5 (choose rule_107).

A5 -> A9 A13 (rule_410).

Now apply rule modifications to have A9 produce the first longer segment, A13 produce the second segment.

Now we need to ensure the terminals match.

Define A9 using rule_427 (modified) to produce:

25 17 14 16 (then A16 part) 15 18 19 17.

A16 will produce "23 22 21 20 24 1". Let's confirm: Use sequence A16 -> A17 A16 (choose A17 => 23). Then recursion A16 -> A17 A16 again -> choose 22 ... Continue until final base: A16 -> 1 (rule_453). So we can generate exactly "23 22 21 20 24 1". To ensure the order as needed: yes.

Thus A9 produces:

25 17 14 16 23 22 21 20 24 1 15 18 19 17.

Now A13 using modified rule_444 (insert 12 4) to produce:

13 3 11 2 A14 12 4 5 3.

Now A14 must generate "10 6 7 9 8 1". Use rule A14 -> A15 A14 recursion repeated. Starting sequence: choose A15=10, then recursion; choose A15=6, recursion; choose 7; 9; 8; then base 1. That yields "10 6 7 9 8 1". Good.

Thus final output string becomes:

1 1 25 17 14 16 23 22 21 20 24 1 15 18 19 17 13 3 11 2 10 6 7 9 8 1 12 4 5 3

Exactly matches target sequence.

Thus we have inserted terminals only: a "17" after 25 in rule_427, and "12 4" after A14 in rule_444. Wait also note that in our derived A13 rule, the order is 13, 3, 11, 2, A14, 12, 4, 5, 3. This matches target final "5 3". The target final segment is "... 5 3" indeed. So fine.

Need to ensure that we inserted "12" and "4" after "A14" before "5". Yes.

Now confirm counts: Are there any other mismatches? Let's double-check the target after "13 3 11 2" appears "10". But according to our A14, first A15 picks "10". But A14 must also generate "10"? A15 can produce 10 (rule_448). Acceptable.

Now ensure that A14's expansions after base 1 produce "10 6 7 9 8 1"? But recursion: A14 -> A15 A14; we can choose A15=10; then A14 (remaining). Next recursion: again choose A15=6; etc. After final recursion we must end with base "1". So we produce "10 6 7 9 8 1". That matches.

Now need to verify again that there is no "14" after 25 before 16 etc. Wait target includes "25 17 14 16". In rule modification we inserted "17" after 25, but we also must ensure that "14" and "16" appear after 17. Yes as rule originally had "25 14 16 ...". So after inserting 17 between 25 and 14, we get "25 17 14 16". That's correct.

Now confirm there is "14" before "16" (the target has "14 16"). Yes.

Now the sequence remains: "25 17 14 16 23 22 21 20 24 1 15 18 19 17". Wait we inserted 17 after 25, but we also need the 17 after the whole block? Actually target includes a 17 after "19"? There are two occurrences: at index #16 (after 19) in the target: "19 17". In current A9 rule, there is a 17 at the very end already. So we have "... 19 17". So with the insertion we get two 17s: one after 25 (makes "25 17 14 16 ...") and one at end of A9 after 19. That matches target which has a 17 after 25 and a 17 after 19. Good.

Now need to make sure there is no extra "17" after the "15"? Actually A9 segment includes "... 15 18 19 17". No 17 after 15, fine.

Now A13's segment includes another "13" at start, matches target where after the A9 segment we have "13". Then "3 11 2". So far all good.

Now the final "5 3" matches target's final two symbols: the target ends "5 3". Indeed final symbol is 3. So everything matches.

Thus only insertions needed: in rule_427 add "17", and in rule_444 add "12" and "4".

Now as per instructions: Only list updated production rules (including rule_<n>:). Provide new lines for each updated rule: rule_427 and rule_444. We have to reprint the full right-hand side with inserted terminals. Should we also give a "..." preceding modifications? Just write them.

Thus:

rule_427: A9 -> 25 17 14 16 A16 15 18 19 17

rule_444: A13 -> 13 3 11 2 A14 12 4 5 3

Check whether rule numbers are exactly as given: rule_427 originally: A9 -> 25 14 16 A16 15 18 19 17. Updated to include 17 after 25. Yes.

rule_444 originally: A13 -> 13 3 11 2 A14 5 3. Updated to include 12 4 before 5.

Make sure to maintain same other spaces; order: "A9 -> 25 17 14 16 A16 15 18 19 17". For rule_444: "A13 -> 13 3 11 2 A14 12 4 5 3". This may be ambiguous: "5" is terminal "5"? Yes.

Now we should give final answer as:

Updated production rules:
rule_427: A9 -> 25 17 14 16 A16 15 18 19 17
rule_444: A13 -> 13 3 11 2 A14 12 4 5 3

Make sure format as required.

Potential nuance: The rule number is 444 already uses A13 -> 13 3 11 2 A14 5 3. We inserted 12 and 4 between A14 and 5. So final rule includes "12 4" as terminals. Could we need to also consider spaces? Should be "12 4" separated.

Check if there is any need to insert terminal "14"? Already there.

Now just output.

Before concluding, double-check any other mismatches: In the target after "15 18 19 17" there is "13" then "3", "11", "2". In rule 444 we have "13 3 11 2 ...". Good.

Now check the ordering: We inserted "12 4" after A14, which appears after the sequence generated by A14 (including "10 6 7 9 8 1"). So the final substring becomes "... 10 6 7 9 8 1 12 4 5 3". The target: "... 10 6 7 9 8 1 12 4 5 3". Exactly matches.

Now check we haven't used any other terminals that wouldn't be present: In the A14 generation we produce "10 6 7 9 8 1". That's okay.

Now verify that the A15 productions include the order we need: They include 6,7,8,9,10. So we can pick appropriate.

Now check that A16 or any other expansions don't inadvertently produce extra terminals that we haven't accounted for. A16 is used only inside A9 to produce the middle sequence. We have A16 generating "23 22 21 20 24 1". Good.

We need to ensure that there is A16 production for 23,22,21,20,24 choices using A17. The rules for A17 are: rule_455: A17 -> 20; rule_456: 21; rule_457: 22; rule_458: 23; rule_459: 24. So all good.

Now check that A58's else produce no extra terminals: we used rule A58 -> A60 A59 (both produce only "1").

Now A60's production: there are many repeated "A60 -> 1" rules (rules 1858-1862). So fine.

A59's production: many "A59 -> 1". So fine.

Now A3 -> A4 (rule 102). Good.

Now A4 -> A5 (choose rule 107). That yields A5 -> A9 A13 (rule 410). Good.

Thus entire string derived matches target.

Now ensure that there are no contradictions with multiple possible expansions generating additional terminals (like other recursion loops). But we are free to choose derivation path. So okay.

Now report only updated rules.

Potential nuance: In rule_444, originally "A13 -> 13 3 11 2 A14 5 3". But we inserted "12 4". Could we also need to insert "5"? Already there. So final rule: "A13 -> 13 3 11 2 A14 12 4 5 3". Yes.

Double-check: Are there any other rules that might also need insertion to produce the "17" after 25? Could we also simply insert "17" at end of rule_427 after 25? Yes.

Now confirm there is no need to add new rules for "12" and "4" maybe as separate terminal productions? They are already present in grammar: terminals 12 and 4 appear in many productions: rule 1953: A66 -> 12; rule 1135: includes 12; rule 1815 includes 12? Actually A57 includes 12. So they are valid terminals.

Thus insertion works.

Now we just output updated production rules accordingly.

One more nuance: Did we need to consider that A14 would produce "10 6 7 9 8 1". The order is 10 first, yes.

Now check that there is any risk that A15's recursion may produce the numbers in reverse order? Actually A14 -> A15 A14 yields first A15 then recursion yields later A15 etc, base 1 at end. So generating 10 6 ... is straightforward.

Now we have minimal insertions? Currently inserted 3 terminals total (one in rule_427, two in rule_444). Could there be any alternate solution with fewer than 3? Possibly we could possibly use some existing rule to produce 12 and 4 without needing insertion. Let's examine.

Could we avoid inserting 12 4 by using A14's recursion to produce them? A14 can't produce 12 or 4 because they are not from A15. However other nonterminals can produce 12 and 4. For example A66 -> 12 (a single 12 terminal). However we cannot incorporate A66 into the expansion unless inserted somewhere else. Could maybe modify A14's rule to include A66 after recursion? Like change A14 -> A15 A14 to allow insertion of a new nonterminal after recursion, but we can only insert terminal symbols, not nonterminals. So can't swap.

Thus need to add 12 and 4 somewhere. Could alternatively modify A14's base case: A14 -> 1 could become 1 12 4? That's allowed: inserting terminals after 1 perhaps. If we modify rule_446: A14 -> 1 into "A14 -> 1 12 4". However that would add "12 4" but would place them after the final 1 in A14, which would be after the sequence "10 6 7 9 8 1" then "12 4". Wait after recursion, base case is 1, then the insertion would produce "12 4" after that 1, before the rest of A13 (which is "5 3"). So we'd get "10 6 7 9 8 1 12 4 5 3" exactly like needed, by insertion in rule_446 (A14 -> 1) adding "12 4". That would require just 2 terminals inserted as before, but not in rule_444. However we can't guarantee we will have other expansions where A14 uses that base rule. Since A14's recursion ends with base rule, we can add "12 4" to that base rule. So we can keep rule_444 unchanged (no insertion) and modify rule_446. Then we need only insert 3 terminals: the 17 after 25, and 12 and 4 after 1 in rule_446. That is also 3 terminals, same count.

But could we reduce to 2? Could we avoid the separate insertion of 12 and 4 by using a rule that already outputs "12 4" after 1 without insertion? For example A14 -> A15 A14 might produce 1 12 4 via additional recursion using A15 that yields 12? But A15 does not include 12. However A14 -> A15 A14 cannot produce 12 4. So need insertion.

Alternatively maybe use A13's alternative production A13 -> A13 A12, combined with A12's expansions to contain "12 4 5 3". But that would require modifications elsewhere. It might be possible to avoid insertion 12 and 4 by using A13 -> A13 A12 and picking A13->some pattern that produces preceding segment, and A12 produce "12 4 5 3". But A12's productions include: A12 -> 1, A12 -> A10 A13, A12 -> A11 A5, A12 -> A12 A12. No production directly generating "12 4" as terminals. However we could have A12-> something that yields "12 4 ..." e.g., A12 -> A10 A13, and perhaps A10 or A13 could produce the needed 12 4, not likely.

Thus minimal insert count is likely 3.

But maybe we can combine insertion of 12 and 4 into the same rule where we add both at once, which we already do either in rule_444 or rule_446. So count stays 2 insertions for that segment (12 and 4). So total of 3 inserted terminals.

Is there any other missing terminal needed? Not upward.

Hence answer is the two updated rules.

Now formatting.

Our output must be:

Updated production rules:
rule_427: A9 -> 25 17 14 16 A16 15 18 19 17
rule_444: A13 -> 13 3 11 2 A14 12 4 5 3

Make sure proper spaces.

Now final.Updated production rules:
rule_427: A9 -> 25 17 14 16 A16 15 18 19 17
rule_444: A13 -> 13 3 11 2 A14 12 4 5 3