(0) Obligation:

Clauses:

p(0).
p(s(X)) :- ','(q(X), ','(!, r)).
p(X) :- r.
q(0).
q(s(X)) :- ','(p(X), ','(!, r)).
q(X) :- r.
r.

Query: p(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

(2) Obligation:

Clauses:

pA(0).
pA(0).
pA(s(0)).
pA(s(s(T10))) :- pA(T10).
pA(s(s(T10))) :- pA(T10).
pA(s(T13)).
pA(s(T16)).
pA(T18).

Query: pA(g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
pA_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)

(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:

PA_IN_G(s(s(T10))) → U1_G(T10, pA_in_g(T10))
PA_IN_G(s(s(T10))) → PA_IN_G(T10)

The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
PA_IN_G(x1)  =  PA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(6) Obligation:

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

PA_IN_G(s(s(T10))) → U1_G(T10, pA_in_g(T10))
PA_IN_G(s(s(T10))) → PA_IN_G(T10)

The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
PA_IN_G(x1)  =  PA_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

PA_IN_G(s(s(T10))) → PA_IN_G(T10)

The TRS R consists of the following rules:

pA_in_g(0) → pA_out_g(0)
pA_in_g(s(0)) → pA_out_g(s(0))
pA_in_g(s(s(T10))) → U1_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U1_g(T10, pA_out_g(T10)) → pA_out_g(s(s(T10)))

The argument filtering Pi contains the following mapping:
pA_in_g(x1)  =  pA_in_g(x1)
0  =  0
pA_out_g(x1)  =  pA_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
PA_IN_G(x1)  =  PA_IN_G(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:

PA_IN_G(s(s(T10))) → PA_IN_G(T10)

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

(11) PiDPToQDPProof (EQUIVALENT 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:

PA_IN_G(s(s(T10))) → PA_IN_G(T10)

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:

  • PA_IN_G(s(s(T10))) → PA_IN_G(T10)
    The graph contains the following edges 1 > 1

(14) YES