(0) Obligation:

Clauses:

p(X, X).
p(f(X), g(Y)) :- ','(p(f(X), f(Z)), p(Z, g(W))).

Queries:

p(g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

p6(T21, T21, X9) :- p11(T21, X9).
p11(g(T35), T35).
p11(f(T38), X41) :- p6(T38, X39, X40).
p1(T4, T4).
p1(f(T7), g(T8)) :- p6(T7, X8, X9).
p1(f(T57), g(T44)) :- p11(T57, X55).

Queries:

p1(g,a).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
p1_in: (b,f)
p6_in: (b,f,f)
p11_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T7), g(T8)) → U3_ga(T7, T8, p6_in_gaa(T7, X8, X9))
p6_in_gaa(T21, T21, X9) → U1_gaa(T21, X9, p11_in_ga(T21, X9))
p11_in_ga(g(T35), T35) → p11_out_ga(g(T35), T35)
p11_in_ga(f(T38), X41) → U2_ga(T38, X41, p6_in_gaa(T38, X39, X40))
U2_ga(T38, X41, p6_out_gaa(T38, X39, X40)) → p11_out_ga(f(T38), X41)
U1_gaa(T21, X9, p11_out_ga(T21, X9)) → p6_out_gaa(T21, T21, X9)
U3_ga(T7, T8, p6_out_gaa(T7, X8, X9)) → p1_out_ga(f(T7), g(T8))
p1_in_ga(f(T57), g(T44)) → U4_ga(T57, T44, p11_in_ga(T57, X55))
U4_ga(T57, T44, p11_out_ga(T57, X55)) → p1_out_ga(f(T57), g(T44))

The argument filtering Pi contains the following mapping:
p1_in_ga(x1, x2)  =  p1_in_ga(x1)
p1_out_ga(x1, x2)  =  p1_out_ga(x1, x2)
f(x1)  =  f(x1)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
p6_in_gaa(x1, x2, x3)  =  p6_in_gaa(x1)
U1_gaa(x1, x2, x3)  =  U1_gaa(x1, x3)
p11_in_ga(x1, x2)  =  p11_in_ga(x1)
g(x1)  =  g
p11_out_ga(x1, x2)  =  p11_out_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
p6_out_gaa(x1, x2, x3)  =  p6_out_gaa(x1, x2)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T7), g(T8)) → U3_ga(T7, T8, p6_in_gaa(T7, X8, X9))
p6_in_gaa(T21, T21, X9) → U1_gaa(T21, X9, p11_in_ga(T21, X9))
p11_in_ga(g(T35), T35) → p11_out_ga(g(T35), T35)
p11_in_ga(f(T38), X41) → U2_ga(T38, X41, p6_in_gaa(T38, X39, X40))
U2_ga(T38, X41, p6_out_gaa(T38, X39, X40)) → p11_out_ga(f(T38), X41)
U1_gaa(T21, X9, p11_out_ga(T21, X9)) → p6_out_gaa(T21, T21, X9)
U3_ga(T7, T8, p6_out_gaa(T7, X8, X9)) → p1_out_ga(f(T7), g(T8))
p1_in_ga(f(T57), g(T44)) → U4_ga(T57, T44, p11_in_ga(T57, X55))
U4_ga(T57, T44, p11_out_ga(T57, X55)) → p1_out_ga(f(T57), g(T44))

The argument filtering Pi contains the following mapping:
p1_in_ga(x1, x2)  =  p1_in_ga(x1)
p1_out_ga(x1, x2)  =  p1_out_ga(x1, x2)
f(x1)  =  f(x1)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
p6_in_gaa(x1, x2, x3)  =  p6_in_gaa(x1)
U1_gaa(x1, x2, x3)  =  U1_gaa(x1, x3)
p11_in_ga(x1, x2)  =  p11_in_ga(x1)
g(x1)  =  g
p11_out_ga(x1, x2)  =  p11_out_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
p6_out_gaa(x1, x2, x3)  =  p6_out_gaa(x1, x2)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)

(5) DependencyPairsProof (EQUIVALENT transformation)

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

P1_IN_GA(f(T7), g(T8)) → U3_GA(T7, T8, p6_in_gaa(T7, X8, X9))
P1_IN_GA(f(T7), g(T8)) → P6_IN_GAA(T7, X8, X9)
P6_IN_GAA(T21, T21, X9) → U1_GAA(T21, X9, p11_in_ga(T21, X9))
P6_IN_GAA(T21, T21, X9) → P11_IN_GA(T21, X9)
P11_IN_GA(f(T38), X41) → U2_GA(T38, X41, p6_in_gaa(T38, X39, X40))
P11_IN_GA(f(T38), X41) → P6_IN_GAA(T38, X39, X40)
P1_IN_GA(f(T57), g(T44)) → U4_GA(T57, T44, p11_in_ga(T57, X55))
P1_IN_GA(f(T57), g(T44)) → P11_IN_GA(T57, X55)

The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T7), g(T8)) → U3_ga(T7, T8, p6_in_gaa(T7, X8, X9))
p6_in_gaa(T21, T21, X9) → U1_gaa(T21, X9, p11_in_ga(T21, X9))
p11_in_ga(g(T35), T35) → p11_out_ga(g(T35), T35)
p11_in_ga(f(T38), X41) → U2_ga(T38, X41, p6_in_gaa(T38, X39, X40))
U2_ga(T38, X41, p6_out_gaa(T38, X39, X40)) → p11_out_ga(f(T38), X41)
U1_gaa(T21, X9, p11_out_ga(T21, X9)) → p6_out_gaa(T21, T21, X9)
U3_ga(T7, T8, p6_out_gaa(T7, X8, X9)) → p1_out_ga(f(T7), g(T8))
p1_in_ga(f(T57), g(T44)) → U4_ga(T57, T44, p11_in_ga(T57, X55))
U4_ga(T57, T44, p11_out_ga(T57, X55)) → p1_out_ga(f(T57), g(T44))

The argument filtering Pi contains the following mapping:
p1_in_ga(x1, x2)  =  p1_in_ga(x1)
p1_out_ga(x1, x2)  =  p1_out_ga(x1, x2)
f(x1)  =  f(x1)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
p6_in_gaa(x1, x2, x3)  =  p6_in_gaa(x1)
U1_gaa(x1, x2, x3)  =  U1_gaa(x1, x3)
p11_in_ga(x1, x2)  =  p11_in_ga(x1)
g(x1)  =  g
p11_out_ga(x1, x2)  =  p11_out_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
p6_out_gaa(x1, x2, x3)  =  p6_out_gaa(x1, x2)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
P1_IN_GA(x1, x2)  =  P1_IN_GA(x1)
U3_GA(x1, x2, x3)  =  U3_GA(x1, x3)
P6_IN_GAA(x1, x2, x3)  =  P6_IN_GAA(x1)
U1_GAA(x1, x2, x3)  =  U1_GAA(x1, x3)
P11_IN_GA(x1, x2)  =  P11_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U4_GA(x1, x2, x3)  =  U4_GA(x1, x3)

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

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P1_IN_GA(f(T7), g(T8)) → U3_GA(T7, T8, p6_in_gaa(T7, X8, X9))
P1_IN_GA(f(T7), g(T8)) → P6_IN_GAA(T7, X8, X9)
P6_IN_GAA(T21, T21, X9) → U1_GAA(T21, X9, p11_in_ga(T21, X9))
P6_IN_GAA(T21, T21, X9) → P11_IN_GA(T21, X9)
P11_IN_GA(f(T38), X41) → U2_GA(T38, X41, p6_in_gaa(T38, X39, X40))
P11_IN_GA(f(T38), X41) → P6_IN_GAA(T38, X39, X40)
P1_IN_GA(f(T57), g(T44)) → U4_GA(T57, T44, p11_in_ga(T57, X55))
P1_IN_GA(f(T57), g(T44)) → P11_IN_GA(T57, X55)

The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T7), g(T8)) → U3_ga(T7, T8, p6_in_gaa(T7, X8, X9))
p6_in_gaa(T21, T21, X9) → U1_gaa(T21, X9, p11_in_ga(T21, X9))
p11_in_ga(g(T35), T35) → p11_out_ga(g(T35), T35)
p11_in_ga(f(T38), X41) → U2_ga(T38, X41, p6_in_gaa(T38, X39, X40))
U2_ga(T38, X41, p6_out_gaa(T38, X39, X40)) → p11_out_ga(f(T38), X41)
U1_gaa(T21, X9, p11_out_ga(T21, X9)) → p6_out_gaa(T21, T21, X9)
U3_ga(T7, T8, p6_out_gaa(T7, X8, X9)) → p1_out_ga(f(T7), g(T8))
p1_in_ga(f(T57), g(T44)) → U4_ga(T57, T44, p11_in_ga(T57, X55))
U4_ga(T57, T44, p11_out_ga(T57, X55)) → p1_out_ga(f(T57), g(T44))

The argument filtering Pi contains the following mapping:
p1_in_ga(x1, x2)  =  p1_in_ga(x1)
p1_out_ga(x1, x2)  =  p1_out_ga(x1, x2)
f(x1)  =  f(x1)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
p6_in_gaa(x1, x2, x3)  =  p6_in_gaa(x1)
U1_gaa(x1, x2, x3)  =  U1_gaa(x1, x3)
p11_in_ga(x1, x2)  =  p11_in_ga(x1)
g(x1)  =  g
p11_out_ga(x1, x2)  =  p11_out_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
p6_out_gaa(x1, x2, x3)  =  p6_out_gaa(x1, x2)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
P1_IN_GA(x1, x2)  =  P1_IN_GA(x1)
U3_GA(x1, x2, x3)  =  U3_GA(x1, x3)
P6_IN_GAA(x1, x2, x3)  =  P6_IN_GAA(x1)
U1_GAA(x1, x2, x3)  =  U1_GAA(x1, x3)
P11_IN_GA(x1, x2)  =  P11_IN_GA(x1)
U2_GA(x1, x2, x3)  =  U2_GA(x1, x3)
U4_GA(x1, x2, x3)  =  U4_GA(x1, x3)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 6 less nodes.

(8) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P6_IN_GAA(T21, T21, X9) → P11_IN_GA(T21, X9)
P11_IN_GA(f(T38), X41) → P6_IN_GAA(T38, X39, X40)

The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T7), g(T8)) → U3_ga(T7, T8, p6_in_gaa(T7, X8, X9))
p6_in_gaa(T21, T21, X9) → U1_gaa(T21, X9, p11_in_ga(T21, X9))
p11_in_ga(g(T35), T35) → p11_out_ga(g(T35), T35)
p11_in_ga(f(T38), X41) → U2_ga(T38, X41, p6_in_gaa(T38, X39, X40))
U2_ga(T38, X41, p6_out_gaa(T38, X39, X40)) → p11_out_ga(f(T38), X41)
U1_gaa(T21, X9, p11_out_ga(T21, X9)) → p6_out_gaa(T21, T21, X9)
U3_ga(T7, T8, p6_out_gaa(T7, X8, X9)) → p1_out_ga(f(T7), g(T8))
p1_in_ga(f(T57), g(T44)) → U4_ga(T57, T44, p11_in_ga(T57, X55))
U4_ga(T57, T44, p11_out_ga(T57, X55)) → p1_out_ga(f(T57), g(T44))

The argument filtering Pi contains the following mapping:
p1_in_ga(x1, x2)  =  p1_in_ga(x1)
p1_out_ga(x1, x2)  =  p1_out_ga(x1, x2)
f(x1)  =  f(x1)
U3_ga(x1, x2, x3)  =  U3_ga(x1, x3)
p6_in_gaa(x1, x2, x3)  =  p6_in_gaa(x1)
U1_gaa(x1, x2, x3)  =  U1_gaa(x1, x3)
p11_in_ga(x1, x2)  =  p11_in_ga(x1)
g(x1)  =  g
p11_out_ga(x1, x2)  =  p11_out_ga(x1)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
p6_out_gaa(x1, x2, x3)  =  p6_out_gaa(x1, x2)
U4_ga(x1, x2, x3)  =  U4_ga(x1, x3)
P6_IN_GAA(x1, x2, x3)  =  P6_IN_GAA(x1)
P11_IN_GA(x1, x2)  =  P11_IN_GA(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

P6_IN_GAA(T21, T21, X9) → P11_IN_GA(T21, X9)
P11_IN_GA(f(T38), X41) → P6_IN_GAA(T38, X39, X40)

R is empty.
The argument filtering Pi contains the following mapping:
f(x1)  =  f(x1)
P6_IN_GAA(x1, x2, x3)  =  P6_IN_GAA(x1)
P11_IN_GA(x1, x2)  =  P11_IN_GA(x1)

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

(11) PiDPToQDPProof (SOUND transformation)

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

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

P6_IN_GAA(T21) → P11_IN_GA(T21)
P11_IN_GA(f(T38)) → P6_IN_GAA(T38)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

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:

  • P11_IN_GA(f(T38)) → P6_IN_GAA(T38)
    The graph contains the following edges 1 > 1

  • P6_IN_GAA(T21) → P11_IN_GA(T21)
    The graph contains the following edges 1 >= 1

(14) YES