YES
proof of /hpcwork/ff862203/termcomp26/benchmarks/ksWKK.pl
# AProVE Commit ID: 23a904c96b029b0a549cde0d0d17dbccf967db59 jckassing 20260626 unpublished dirty


Left Termination of the query pattern

div(g,g,a)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) PiDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) PiDP
        (10) PiDPToQDPProof [SOUND, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) PiDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) PiDP
        (17) PiDPToQDPProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) PiDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) PiDP
        (24) PiDPToQDPProof [SOUND, 0 ms]
        (25) QDP
        (26) QDPOrderProof [EQUIVALENT, 5 ms]
        (27) QDP
        (28) DependencyGraphProof [EQUIVALENT, 0 ms]
        (29) TRUE


----------------------------------------

(0)
Obligation:
Clauses:

div(X, s(Y), Z) :- div_s(X, Y, Z).
div_s(0, Y, 0).
div_s(s(X), Y, 0) :- lss(X, Y).
div_s(s(X), Y, s(Z)) :- ','(sub(X, Y, R), div_s(R, Y, Z)).
lss(s(X), s(Y)) :- lss(X, Y).
lss(0, s(Y)).
sub(s(X), s(Y), Z) :- sub(X, Y, Z).
sub(X, 0, X).


Query: div(g,g,a)
----------------------------------------

(1) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

div_in_3: (b,b,f)

div_s_in_3: (b,b,f)

lss_in_2: (b,b)

sub_in_3: (b,b,f)

Transforming Prolog into the following Term Rewriting System:

Pi-finite rewrite system:
The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



----------------------------------------

(2)
Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)



----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

   DIV_IN_GGA(X, s(Y), Z) -> U1_GGA(X, Y, Z, div_s_in_gga(X, Y, Z))
   DIV_IN_GGA(X, s(Y), Z) -> DIV_S_IN_GGA(X, Y, Z)
   DIV_S_IN_GGA(s(X), Y, 0) -> U2_GGA(X, Y, lss_in_gg(X, Y))
   DIV_S_IN_GGA(s(X), Y, 0) -> LSS_IN_GG(X, Y)
   LSS_IN_GG(s(X), s(Y)) -> U5_GG(X, Y, lss_in_gg(X, Y))
   LSS_IN_GG(s(X), s(Y)) -> LSS_IN_GG(X, Y)
   DIV_S_IN_GGA(s(X), Y, s(Z)) -> U3_GGA(X, Y, Z, sub_in_gga(X, Y, R))
   DIV_S_IN_GGA(s(X), Y, s(Z)) -> SUB_IN_GGA(X, Y, R)
   SUB_IN_GGA(s(X), s(Y), Z) -> U6_GGA(X, Y, Z, sub_in_gga(X, Y, Z))
   SUB_IN_GGA(s(X), s(Y), Z) -> SUB_IN_GGA(X, Y, Z)
   U3_GGA(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_GGA(X, Y, Z, div_s_in_gga(R, Y, Z))
   U3_GGA(X, Y, Z, sub_out_gga(X, Y, R)) -> DIV_S_IN_GGA(R, Y, Z)

The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)

DIV_IN_GGA(x1, x2, x3)  =  DIV_IN_GGA(x1, x2)

U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x4)

DIV_S_IN_GGA(x1, x2, x3)  =  DIV_S_IN_GGA(x1, x2)

U2_GGA(x1, x2, x3)  =  U2_GGA(x3)

LSS_IN_GG(x1, x2)  =  LSS_IN_GG(x1, x2)

U5_GG(x1, x2, x3)  =  U5_GG(x3)

U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x2, x4)

SUB_IN_GGA(x1, x2, x3)  =  SUB_IN_GGA(x1, x2)

U6_GGA(x1, x2, x3, x4)  =  U6_GGA(x4)

U4_GGA(x1, x2, x3, x4)  =  U4_GGA(x4)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(4)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   DIV_IN_GGA(X, s(Y), Z) -> U1_GGA(X, Y, Z, div_s_in_gga(X, Y, Z))
   DIV_IN_GGA(X, s(Y), Z) -> DIV_S_IN_GGA(X, Y, Z)
   DIV_S_IN_GGA(s(X), Y, 0) -> U2_GGA(X, Y, lss_in_gg(X, Y))
   DIV_S_IN_GGA(s(X), Y, 0) -> LSS_IN_GG(X, Y)
   LSS_IN_GG(s(X), s(Y)) -> U5_GG(X, Y, lss_in_gg(X, Y))
   LSS_IN_GG(s(X), s(Y)) -> LSS_IN_GG(X, Y)
   DIV_S_IN_GGA(s(X), Y, s(Z)) -> U3_GGA(X, Y, Z, sub_in_gga(X, Y, R))
   DIV_S_IN_GGA(s(X), Y, s(Z)) -> SUB_IN_GGA(X, Y, R)
   SUB_IN_GGA(s(X), s(Y), Z) -> U6_GGA(X, Y, Z, sub_in_gga(X, Y, Z))
   SUB_IN_GGA(s(X), s(Y), Z) -> SUB_IN_GGA(X, Y, Z)
   U3_GGA(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_GGA(X, Y, Z, div_s_in_gga(R, Y, Z))
   U3_GGA(X, Y, Z, sub_out_gga(X, Y, R)) -> DIV_S_IN_GGA(R, Y, Z)

The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)

DIV_IN_GGA(x1, x2, x3)  =  DIV_IN_GGA(x1, x2)

U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x4)

DIV_S_IN_GGA(x1, x2, x3)  =  DIV_S_IN_GGA(x1, x2)

U2_GGA(x1, x2, x3)  =  U2_GGA(x3)

LSS_IN_GG(x1, x2)  =  LSS_IN_GG(x1, x2)

U5_GG(x1, x2, x3)  =  U5_GG(x3)

U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x2, x4)

SUB_IN_GGA(x1, x2, x3)  =  SUB_IN_GGA(x1, x2)

U6_GGA(x1, x2, x3, x4)  =  U6_GGA(x4)

U4_GGA(x1, x2, x3, x4)  =  U4_GGA(x4)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 8 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

----------------------------------------

(7)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SUB_IN_GGA(s(X), s(Y), Z) -> SUB_IN_GGA(X, Y, Z)

The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)

SUB_IN_GGA(x1, x2, x3)  =  SUB_IN_GGA(x1, x2)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(8) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
----------------------------------------

(9)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SUB_IN_GGA(s(X), s(Y), Z) -> SUB_IN_GGA(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)

SUB_IN_GGA(x1, x2, x3)  =  SUB_IN_GGA(x1, x2)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(10) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   SUB_IN_GGA(s(X), s(Y)) -> SUB_IN_GGA(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
----------------------------------------

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*SUB_IN_GGA(s(X), s(Y)) -> SUB_IN_GGA(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   LSS_IN_GG(s(X), s(Y)) -> LSS_IN_GG(X, Y)

The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)

LSS_IN_GG(x1, x2)  =  LSS_IN_GG(x1, x2)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(15) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
----------------------------------------

(16)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   LSS_IN_GG(s(X), s(Y)) -> LSS_IN_GG(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
----------------------------------------

(17) PiDPToQDPProof (EQUIVALENT)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LSS_IN_GG(s(X), s(Y)) -> LSS_IN_GG(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
----------------------------------------

(19) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*LSS_IN_GG(s(X), s(Y)) -> LSS_IN_GG(X, Y)
The graph contains the following edges 1 > 1, 2 > 2


----------------------------------------

(20)
YES

----------------------------------------

(21)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   DIV_S_IN_GGA(s(X), Y, s(Z)) -> U3_GGA(X, Y, Z, sub_in_gga(X, Y, R))
   U3_GGA(X, Y, Z, sub_out_gga(X, Y, R)) -> DIV_S_IN_GGA(R, Y, Z)

The TRS R consists of the following rules:

   div_in_gga(X, s(Y), Z) -> U1_gga(X, Y, Z, div_s_in_gga(X, Y, Z))
   div_s_in_gga(0, Y, 0) -> div_s_out_gga(0, Y, 0)
   div_s_in_gga(s(X), Y, 0) -> U2_gga(X, Y, lss_in_gg(X, Y))
   lss_in_gg(s(X), s(Y)) -> U5_gg(X, Y, lss_in_gg(X, Y))
   lss_in_gg(0, s(Y)) -> lss_out_gg(0, s(Y))
   U5_gg(X, Y, lss_out_gg(X, Y)) -> lss_out_gg(s(X), s(Y))
   U2_gga(X, Y, lss_out_gg(X, Y)) -> div_s_out_gga(s(X), Y, 0)
   div_s_in_gga(s(X), Y, s(Z)) -> U3_gga(X, Y, Z, sub_in_gga(X, Y, R))
   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)
   U3_gga(X, Y, Z, sub_out_gga(X, Y, R)) -> U4_gga(X, Y, Z, div_s_in_gga(R, Y, Z))
   U4_gga(X, Y, Z, div_s_out_gga(R, Y, Z)) -> div_s_out_gga(s(X), Y, s(Z))
   U1_gga(X, Y, Z, div_s_out_gga(X, Y, Z)) -> div_out_gga(X, s(Y), Z)

The argument filtering Pi contains the following mapping:
div_in_gga(x1, x2, x3)  =  div_in_gga(x1, x2)

s(x1)  =  s(x1)

U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)

div_s_in_gga(x1, x2, x3)  =  div_s_in_gga(x1, x2)

0  =  0

div_s_out_gga(x1, x2, x3)  =  div_s_out_gga(x3)

U2_gga(x1, x2, x3)  =  U2_gga(x3)

lss_in_gg(x1, x2)  =  lss_in_gg(x1, x2)

U5_gg(x1, x2, x3)  =  U5_gg(x3)

lss_out_gg(x1, x2)  =  lss_out_gg

U3_gga(x1, x2, x3, x4)  =  U3_gga(x2, x4)

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

U4_gga(x1, x2, x3, x4)  =  U4_gga(x4)

div_out_gga(x1, x2, x3)  =  div_out_gga(x3)

DIV_S_IN_GGA(x1, x2, x3)  =  DIV_S_IN_GGA(x1, x2)

U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x2, x4)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(22) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
----------------------------------------

(23)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   DIV_S_IN_GGA(s(X), Y, s(Z)) -> U3_GGA(X, Y, Z, sub_in_gga(X, Y, R))
   U3_GGA(X, Y, Z, sub_out_gga(X, Y, R)) -> DIV_S_IN_GGA(R, Y, Z)

The TRS R consists of the following rules:

   sub_in_gga(s(X), s(Y), Z) -> U6_gga(X, Y, Z, sub_in_gga(X, Y, Z))
   sub_in_gga(X, 0, X) -> sub_out_gga(X, 0, X)
   U6_gga(X, Y, Z, sub_out_gga(X, Y, Z)) -> sub_out_gga(s(X), s(Y), Z)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)

0  =  0

sub_in_gga(x1, x2, x3)  =  sub_in_gga(x1, x2)

U6_gga(x1, x2, x3, x4)  =  U6_gga(x4)

sub_out_gga(x1, x2, x3)  =  sub_out_gga(x3)

DIV_S_IN_GGA(x1, x2, x3)  =  DIV_S_IN_GGA(x1, x2)

U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x2, x4)


We have to consider all (P,R,Pi)-chains
----------------------------------------

(24) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV_S_IN_GGA(s(X), Y) -> U3_GGA(Y, sub_in_gga(X, Y))
   U3_GGA(Y, sub_out_gga(R)) -> DIV_S_IN_GGA(R, Y)

The TRS R consists of the following rules:

   sub_in_gga(s(X), s(Y)) -> U6_gga(sub_in_gga(X, Y))
   sub_in_gga(X, 0) -> sub_out_gga(X)
   U6_gga(sub_out_gga(Z)) -> sub_out_gga(Z)

The set Q consists of the following terms:

   sub_in_gga(x0, x1)
   U6_gga(x0)

We have to consider all (P,Q,R)-chains.
----------------------------------------

(26) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   U3_GGA(Y, sub_out_gga(R)) -> DIV_S_IN_GGA(R, Y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
DIV_S_IN_GGA(x1, x2)  =  x1

s(x1)  =  s(x1)

U3_GGA(x1, x2)  =  x2

sub_in_gga(x1, x2)  =  sub_in_gga(x1)

sub_out_gga(x1)  =  sub_out_gga(x1)

U6_gga(x1)  =  x1

0  =  0


Recursive path order with status [RPO].
Quasi-Precedence: [s_1, sub_in_gga_1] > sub_out_gga_1

Status: s_1: multiset status
sub_in_gga_1: multiset status
sub_out_gga_1: multiset status
0: multiset status


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   sub_in_gga(s(X), s(Y)) -> U6_gga(sub_in_gga(X, Y))
   sub_in_gga(X, 0) -> sub_out_gga(X)
   U6_gga(sub_out_gga(Z)) -> sub_out_gga(Z)


----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV_S_IN_GGA(s(X), Y) -> U3_GGA(Y, sub_in_gga(X, Y))

The TRS R consists of the following rules:

   sub_in_gga(s(X), s(Y)) -> U6_gga(sub_in_gga(X, Y))
   sub_in_gga(X, 0) -> sub_out_gga(X)
   U6_gga(sub_out_gga(Z)) -> sub_out_gga(Z)

The set Q consists of the following terms:

   sub_in_gga(x0, x1)
   U6_gga(x0)

We have to consider all (P,Q,R)-chains.
----------------------------------------

(28) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(29)
TRUE
