(0) Obligation:

Clauses:

q(X) :- p(X, 0).
p(0, X1).
p(s(X), Y) :- p(X, s(Y)).

Query: q(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

(2) Obligation:

Clauses:

pA(0, T35).
pA(s(T40), T41) :- pA(T40, s(T41)).
qB(0).
qB(s(0)).
qB(s(s(0))).
qB(s(s(s(0)))).
qB(s(s(s(s(0))))).
qB(s(s(s(s(s(0)))))).
qB(s(s(s(s(s(s(0))))))).
qB(s(s(s(s(s(s(s(0)))))))).
qB(s(s(s(s(s(s(s(s(T27))))))))) :- pA(T27, s(s(s(s(s(s(s(0)))))))).

Query: qB(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:
qB_in: (b)
pA_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

qB_in_g(0) → qB_out_g(0)
qB_in_g(s(0)) → qB_out_g(s(0))
qB_in_g(s(s(0))) → qB_out_g(s(s(0)))
qB_in_g(s(s(s(0)))) → qB_out_g(s(s(s(0))))
qB_in_g(s(s(s(s(0))))) → qB_out_g(s(s(s(s(0)))))
qB_in_g(s(s(s(s(s(0)))))) → qB_out_g(s(s(s(s(s(0))))))
qB_in_g(s(s(s(s(s(s(0))))))) → qB_out_g(s(s(s(s(s(s(0)))))))
qB_in_g(s(s(s(s(s(s(s(0)))))))) → qB_out_g(s(s(s(s(s(s(s(0))))))))
qB_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U2_g(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pA_in_gg(0, T35) → pA_out_gg(0, T35)
pA_in_gg(s(T40), T41) → U1_gg(T40, T41, pA_in_gg(T40, s(T41)))
U1_gg(T40, T41, pA_out_gg(T40, s(T41))) → pA_out_gg(s(T40), T41)
U2_g(T27, pA_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qB_out_g(s(s(s(s(s(s(s(s(T27)))))))))

The argument filtering Pi contains the following mapping:
qB_in_g(x1)  =  qB_in_g(x1)
0  =  0
qB_out_g(x1)  =  qB_out_g
s(x1)  =  s(x1)
U2_g(x1, x2)  =  U2_g(x2)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
pA_out_gg(x1, x2)  =  pA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(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:

qB_in_g(0) → qB_out_g(0)
qB_in_g(s(0)) → qB_out_g(s(0))
qB_in_g(s(s(0))) → qB_out_g(s(s(0)))
qB_in_g(s(s(s(0)))) → qB_out_g(s(s(s(0))))
qB_in_g(s(s(s(s(0))))) → qB_out_g(s(s(s(s(0)))))
qB_in_g(s(s(s(s(s(0)))))) → qB_out_g(s(s(s(s(s(0))))))
qB_in_g(s(s(s(s(s(s(0))))))) → qB_out_g(s(s(s(s(s(s(0)))))))
qB_in_g(s(s(s(s(s(s(s(0)))))))) → qB_out_g(s(s(s(s(s(s(s(0))))))))
qB_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U2_g(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pA_in_gg(0, T35) → pA_out_gg(0, T35)
pA_in_gg(s(T40), T41) → U1_gg(T40, T41, pA_in_gg(T40, s(T41)))
U1_gg(T40, T41, pA_out_gg(T40, s(T41))) → pA_out_gg(s(T40), T41)
U2_g(T27, pA_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qB_out_g(s(s(s(s(s(s(s(s(T27)))))))))

The argument filtering Pi contains the following mapping:
qB_in_g(x1)  =  qB_in_g(x1)
0  =  0
qB_out_g(x1)  =  qB_out_g
s(x1)  =  s(x1)
U2_g(x1, x2)  =  U2_g(x2)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
pA_out_gg(x1, x2)  =  pA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(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:

QB_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → U2_G(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
QB_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → PA_IN_GG(T27, s(s(s(s(s(s(s(0))))))))
PA_IN_GG(s(T40), T41) → U1_GG(T40, T41, pA_in_gg(T40, s(T41)))
PA_IN_GG(s(T40), T41) → PA_IN_GG(T40, s(T41))

The TRS R consists of the following rules:

qB_in_g(0) → qB_out_g(0)
qB_in_g(s(0)) → qB_out_g(s(0))
qB_in_g(s(s(0))) → qB_out_g(s(s(0)))
qB_in_g(s(s(s(0)))) → qB_out_g(s(s(s(0))))
qB_in_g(s(s(s(s(0))))) → qB_out_g(s(s(s(s(0)))))
qB_in_g(s(s(s(s(s(0)))))) → qB_out_g(s(s(s(s(s(0))))))
qB_in_g(s(s(s(s(s(s(0))))))) → qB_out_g(s(s(s(s(s(s(0)))))))
qB_in_g(s(s(s(s(s(s(s(0)))))))) → qB_out_g(s(s(s(s(s(s(s(0))))))))
qB_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U2_g(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pA_in_gg(0, T35) → pA_out_gg(0, T35)
pA_in_gg(s(T40), T41) → U1_gg(T40, T41, pA_in_gg(T40, s(T41)))
U1_gg(T40, T41, pA_out_gg(T40, s(T41))) → pA_out_gg(s(T40), T41)
U2_g(T27, pA_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qB_out_g(s(s(s(s(s(s(s(s(T27)))))))))

The argument filtering Pi contains the following mapping:
qB_in_g(x1)  =  qB_in_g(x1)
0  =  0
qB_out_g(x1)  =  qB_out_g
s(x1)  =  s(x1)
U2_g(x1, x2)  =  U2_g(x2)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
pA_out_gg(x1, x2)  =  pA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
QB_IN_G(x1)  =  QB_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x2)
PA_IN_GG(x1, x2)  =  PA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(x3)

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

(6) Obligation:

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

QB_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → U2_G(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
QB_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → PA_IN_GG(T27, s(s(s(s(s(s(s(0))))))))
PA_IN_GG(s(T40), T41) → U1_GG(T40, T41, pA_in_gg(T40, s(T41)))
PA_IN_GG(s(T40), T41) → PA_IN_GG(T40, s(T41))

The TRS R consists of the following rules:

qB_in_g(0) → qB_out_g(0)
qB_in_g(s(0)) → qB_out_g(s(0))
qB_in_g(s(s(0))) → qB_out_g(s(s(0)))
qB_in_g(s(s(s(0)))) → qB_out_g(s(s(s(0))))
qB_in_g(s(s(s(s(0))))) → qB_out_g(s(s(s(s(0)))))
qB_in_g(s(s(s(s(s(0)))))) → qB_out_g(s(s(s(s(s(0))))))
qB_in_g(s(s(s(s(s(s(0))))))) → qB_out_g(s(s(s(s(s(s(0)))))))
qB_in_g(s(s(s(s(s(s(s(0)))))))) → qB_out_g(s(s(s(s(s(s(s(0))))))))
qB_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U2_g(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pA_in_gg(0, T35) → pA_out_gg(0, T35)
pA_in_gg(s(T40), T41) → U1_gg(T40, T41, pA_in_gg(T40, s(T41)))
U1_gg(T40, T41, pA_out_gg(T40, s(T41))) → pA_out_gg(s(T40), T41)
U2_g(T27, pA_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qB_out_g(s(s(s(s(s(s(s(s(T27)))))))))

The argument filtering Pi contains the following mapping:
qB_in_g(x1)  =  qB_in_g(x1)
0  =  0
qB_out_g(x1)  =  qB_out_g
s(x1)  =  s(x1)
U2_g(x1, x2)  =  U2_g(x2)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
pA_out_gg(x1, x2)  =  pA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
QB_IN_G(x1)  =  QB_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x2)
PA_IN_GG(x1, x2)  =  PA_IN_GG(x1, x2)
U1_GG(x1, x2, x3)  =  U1_GG(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 3 less nodes.

(8) Obligation:

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

PA_IN_GG(s(T40), T41) → PA_IN_GG(T40, s(T41))

The TRS R consists of the following rules:

qB_in_g(0) → qB_out_g(0)
qB_in_g(s(0)) → qB_out_g(s(0))
qB_in_g(s(s(0))) → qB_out_g(s(s(0)))
qB_in_g(s(s(s(0)))) → qB_out_g(s(s(s(0))))
qB_in_g(s(s(s(s(0))))) → qB_out_g(s(s(s(s(0)))))
qB_in_g(s(s(s(s(s(0)))))) → qB_out_g(s(s(s(s(s(0))))))
qB_in_g(s(s(s(s(s(s(0))))))) → qB_out_g(s(s(s(s(s(s(0)))))))
qB_in_g(s(s(s(s(s(s(s(0)))))))) → qB_out_g(s(s(s(s(s(s(s(0))))))))
qB_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U2_g(T27, pA_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pA_in_gg(0, T35) → pA_out_gg(0, T35)
pA_in_gg(s(T40), T41) → U1_gg(T40, T41, pA_in_gg(T40, s(T41)))
U1_gg(T40, T41, pA_out_gg(T40, s(T41))) → pA_out_gg(s(T40), T41)
U2_g(T27, pA_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qB_out_g(s(s(s(s(s(s(s(s(T27)))))))))

The argument filtering Pi contains the following mapping:
qB_in_g(x1)  =  qB_in_g(x1)
0  =  0
qB_out_g(x1)  =  qB_out_g
s(x1)  =  s(x1)
U2_g(x1, x2)  =  U2_g(x2)
pA_in_gg(x1, x2)  =  pA_in_gg(x1, x2)
pA_out_gg(x1, x2)  =  pA_out_gg
U1_gg(x1, x2, x3)  =  U1_gg(x3)
PA_IN_GG(x1, x2)  =  PA_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:

PA_IN_GG(s(T40), T41) → PA_IN_GG(T40, s(T41))

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_GG(s(T40), T41) → PA_IN_GG(T40, s(T41))

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_GG(s(T40), T41) → PA_IN_GG(T40, s(T41))
    The graph contains the following edges 1 > 1

(14) YES