(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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

(2) 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, pB_in_g(T10))
pB_in_g(T10) → U2_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U2_g(T10, pA_out_g(T10)) → U3_g(T10, pC_in_)
pC_in_pC_out_
U3_g(T10, pC_out_) → pB_out_g(T10)
U1_g(T10, pB_out_g(T10)) → pA_out_g(s(s(T10)))

Pi is empty.

(3) 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, pB_in_g(T10))
PA_IN_G(s(s(T10))) → PB_IN_G(T10)
PB_IN_G(T10) → U2_G(T10, pA_in_g(T10))
PB_IN_G(T10) → PA_IN_G(T10)
U2_G(T10, pA_out_g(T10)) → U3_G(T10, pC_in_)
U2_G(T10, pA_out_g(T10)) → PC_IN_

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, pB_in_g(T10))
pB_in_g(T10) → U2_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U2_g(T10, pA_out_g(T10)) → U3_g(T10, pC_in_)
pC_in_pC_out_
U3_g(T10, pC_out_) → pB_out_g(T10)
U1_g(T10, pB_out_g(T10)) → pA_out_g(s(s(T10)))

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

(4) Obligation:

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

PA_IN_G(s(s(T10))) → U1_G(T10, pB_in_g(T10))
PA_IN_G(s(s(T10))) → PB_IN_G(T10)
PB_IN_G(T10) → U2_G(T10, pA_in_g(T10))
PB_IN_G(T10) → PA_IN_G(T10)
U2_G(T10, pA_out_g(T10)) → U3_G(T10, pC_in_)
U2_G(T10, pA_out_g(T10)) → PC_IN_

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, pB_in_g(T10))
pB_in_g(T10) → U2_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U2_g(T10, pA_out_g(T10)) → U3_g(T10, pC_in_)
pC_in_pC_out_
U3_g(T10, pC_out_) → pB_out_g(T10)
U1_g(T10, pB_out_g(T10)) → pA_out_g(s(s(T10)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

PA_IN_G(s(s(T10))) → PB_IN_G(T10)
PB_IN_G(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, pB_in_g(T10))
pB_in_g(T10) → U2_g(T10, pA_in_g(T10))
pA_in_g(s(T13)) → pA_out_g(s(T13))
pA_in_g(T18) → pA_out_g(T18)
U2_g(T10, pA_out_g(T10)) → U3_g(T10, pC_in_)
pC_in_pC_out_
U3_g(T10, pC_out_) → pB_out_g(T10)
U1_g(T10, pB_out_g(T10)) → pA_out_g(s(s(T10)))

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

PA_IN_G(s(s(T10))) → PB_IN_G(T10)
PB_IN_G(T10) → PA_IN_G(T10)

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

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

(10) Obligation:

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

PA_IN_G(s(s(T10))) → PB_IN_G(T10)
PB_IN_G(T10) → PA_IN_G(T10)

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

(11) 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:

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

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

(12) YES