Left Termination of the query pattern p_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
↳3 PrologToPiTRSProof (⇐)
↳4 PiTRS
↳5 DependencyPairsProof (⇔)
↳6 PiDP
↳7 DependencyGraphProof (⇔)
↳8 PiDP
↳9 UsableRulesProof (⇔)
↳10 PiDP
↳11 PiDPToQDPProof (⇐)
↳12 QDP
↳13 QDPSizeChangeProof (⇔)
↳14 YES

(0) Obligation:

Clauses:

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

Queries:

p(g,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

p1(T4, T4).
p1(f(T22), g(T9)) :- p1(T22, g(T9)).
p1(f(T45), g(T32)) :- p1(T45, g(T32)).

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) (b,b)
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(T22), g(T9)) → U1_ga(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(T4, T4) → p1_out_gg(T4, T4)
p1_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, p1_in_gg(T45, g(T32)))
U2_gg(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_gg(f(T45), g(T32))
U1_gg(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_gg(f(T22), g(T9))
U1_ga(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_ga(f(T22), g(T9))
p1_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, p1_in_gg(T45, g(T32)))
U2_ga(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_ga(f(T45), g(T32))

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)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
p1_in_gg(x1, x2)  =  p1_in_gg(x1, x2)
g(x1)  =  g
p1_out_gg(x1, x2)  =  p1_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_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(T22), g(T9)) → U1_ga(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(T4, T4) → p1_out_gg(T4, T4)
p1_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, p1_in_gg(T45, g(T32)))
U2_gg(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_gg(f(T45), g(T32))
U1_gg(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_gg(f(T22), g(T9))
U1_ga(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_ga(f(T22), g(T9))
p1_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, p1_in_gg(T45, g(T32)))
U2_ga(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_ga(f(T45), g(T32))

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)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
p1_in_gg(x1, x2)  =  p1_in_gg(x1, x2)
g(x1)  =  g
p1_out_gg(x1, x2)  =  p1_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_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(T22), g(T9)) → U1_GA(T22, T9, p1_in_gg(T22, g(T9)))
P1_IN_GA(f(T22), g(T9)) → P1_IN_GG(T22, g(T9))
P1_IN_GG(f(T22), g(T9)) → U1_GG(T22, T9, p1_in_gg(T22, g(T9)))
P1_IN_GG(f(T22), g(T9)) → P1_IN_GG(T22, g(T9))
P1_IN_GG(f(T45), g(T32)) → U2_GG(T45, T32, p1_in_gg(T45, g(T32)))
P1_IN_GA(f(T45), g(T32)) → U2_GA(T45, T32, p1_in_gg(T45, g(T32)))

The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(T4, T4) → p1_out_gg(T4, T4)
p1_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, p1_in_gg(T45, g(T32)))
U2_gg(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_gg(f(T45), g(T32))
U1_gg(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_gg(f(T22), g(T9))
U1_ga(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_ga(f(T22), g(T9))
p1_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, p1_in_gg(T45, g(T32)))
U2_ga(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_ga(f(T45), g(T32))

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)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
p1_in_gg(x1, x2)  =  p1_in_gg(x1, x2)
g(x1)  =  g
p1_out_gg(x1, x2)  =  p1_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
P1_IN_GA(x1, x2)  =  P1_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
P1_IN_GG(x1, x2)  =  P1_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x3)
U2_GG(x1, x2, x3)  =  U2_GG(x1, x3)
U2_GA(x1, x2, x3)  =  U2_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(T22), g(T9)) → U1_GA(T22, T9, p1_in_gg(T22, g(T9)))
P1_IN_GA(f(T22), g(T9)) → P1_IN_GG(T22, g(T9))
P1_IN_GG(f(T22), g(T9)) → U1_GG(T22, T9, p1_in_gg(T22, g(T9)))
P1_IN_GG(f(T22), g(T9)) → P1_IN_GG(T22, g(T9))
P1_IN_GG(f(T45), g(T32)) → U2_GG(T45, T32, p1_in_gg(T45, g(T32)))
P1_IN_GA(f(T45), g(T32)) → U2_GA(T45, T32, p1_in_gg(T45, g(T32)))

The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(T4, T4) → p1_out_gg(T4, T4)
p1_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, p1_in_gg(T45, g(T32)))
U2_gg(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_gg(f(T45), g(T32))
U1_gg(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_gg(f(T22), g(T9))
U1_ga(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_ga(f(T22), g(T9))
p1_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, p1_in_gg(T45, g(T32)))
U2_ga(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_ga(f(T45), g(T32))

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)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
p1_in_gg(x1, x2)  =  p1_in_gg(x1, x2)
g(x1)  =  g
p1_out_gg(x1, x2)  =  p1_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
P1_IN_GA(x1, x2)  =  P1_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
P1_IN_GG(x1, x2)  =  P1_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x1, x3)
U2_GG(x1, x2, x3)  =  U2_GG(x1, x3)
U2_GA(x1, x2, x3)  =  U2_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 5 less nodes.

(8) Obligation:

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

P1_IN_GG(f(T22), g(T9)) → P1_IN_GG(T22, g(T9))

The TRS R consists of the following rules:

p1_in_ga(T4, T4) → p1_out_ga(T4, T4)
p1_in_ga(f(T22), g(T9)) → U1_ga(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(T4, T4) → p1_out_gg(T4, T4)
p1_in_gg(f(T22), g(T9)) → U1_gg(T22, T9, p1_in_gg(T22, g(T9)))
p1_in_gg(f(T45), g(T32)) → U2_gg(T45, T32, p1_in_gg(T45, g(T32)))
U2_gg(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_gg(f(T45), g(T32))
U1_gg(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_gg(f(T22), g(T9))
U1_ga(T22, T9, p1_out_gg(T22, g(T9))) → p1_out_ga(f(T22), g(T9))
p1_in_ga(f(T45), g(T32)) → U2_ga(T45, T32, p1_in_gg(T45, g(T32)))
U2_ga(T45, T32, p1_out_gg(T45, g(T32))) → p1_out_ga(f(T45), g(T32))

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)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
p1_in_gg(x1, x2)  =  p1_in_gg(x1, x2)
g(x1)  =  g
p1_out_gg(x1, x2)  =  p1_out_gg(x1, x2)
U1_gg(x1, x2, x3)  =  U1_gg(x1, x3)
U2_gg(x1, x2, x3)  =  U2_gg(x1, x3)
U2_ga(x1, x2, x3)  =  U2_ga(x1, x3)
P1_IN_GG(x1, x2)  =  P1_IN_GG(x1, x2)

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:

P1_IN_GG(f(T22), g(T9)) → P1_IN_GG(T22, g(T9))

R is empty.
The argument filtering Pi contains the following mapping:
f(x1)  =  f(x1)
g(x1)  =  g
P1_IN_GG(x1, x2)  =  P1_IN_GG(x1, x2)

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:

P1_IN_GG(f(T22), g) → P1_IN_GG(T22, g)

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:

  • P1_IN_GG(f(T22), g) → P1_IN_GG(T22, g)
    The graph contains the following edges 1 > 1, 2 >= 2

(14) YES