(0) Obligation:

Clauses:

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

Query: q(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:

qA_in_g(0) → qA_out_g(0)
qA_in_g(s(0)) → qA_out_g(s(0))
qA_in_g(s(s(0))) → qA_out_g(s(s(0)))
qA_in_g(s(s(s(0)))) → qA_out_g(s(s(s(0))))
qA_in_g(s(s(s(s(0))))) → qA_out_g(s(s(s(s(0)))))
qA_in_g(s(s(s(s(s(0)))))) → qA_out_g(s(s(s(s(s(0))))))
qA_in_g(s(s(s(s(s(s(0))))))) → qA_out_g(s(s(s(s(s(s(0)))))))
qA_in_g(s(s(s(s(s(s(s(0)))))))) → qA_out_g(s(s(s(s(s(s(s(0))))))))
qA_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U1_g(T27, pB_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pB_in_gg(0, T35) → pB_out_gg(0, T35)
pB_in_gg(s(T40), T41) → U2_gg(T40, T41, pB_in_gg(T40, s(T41)))
U2_gg(T40, T41, pB_out_gg(T40, s(T41))) → pB_out_gg(s(T40), T41)
U1_g(T27, pB_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qA_out_g(s(s(s(s(s(s(s(s(T27)))))))))

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:

QA_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → U1_G(T27, pB_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
QA_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → PB_IN_GG(T27, s(s(s(s(s(s(s(0))))))))
PB_IN_GG(s(T40), T41) → U2_GG(T40, T41, pB_in_gg(T40, s(T41)))
PB_IN_GG(s(T40), T41) → PB_IN_GG(T40, s(T41))

The TRS R consists of the following rules:

qA_in_g(0) → qA_out_g(0)
qA_in_g(s(0)) → qA_out_g(s(0))
qA_in_g(s(s(0))) → qA_out_g(s(s(0)))
qA_in_g(s(s(s(0)))) → qA_out_g(s(s(s(0))))
qA_in_g(s(s(s(s(0))))) → qA_out_g(s(s(s(s(0)))))
qA_in_g(s(s(s(s(s(0)))))) → qA_out_g(s(s(s(s(s(0))))))
qA_in_g(s(s(s(s(s(s(0))))))) → qA_out_g(s(s(s(s(s(s(0)))))))
qA_in_g(s(s(s(s(s(s(s(0)))))))) → qA_out_g(s(s(s(s(s(s(s(0))))))))
qA_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U1_g(T27, pB_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pB_in_gg(0, T35) → pB_out_gg(0, T35)
pB_in_gg(s(T40), T41) → U2_gg(T40, T41, pB_in_gg(T40, s(T41)))
U2_gg(T40, T41, pB_out_gg(T40, s(T41))) → pB_out_gg(s(T40), T41)
U1_g(T27, pB_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qA_out_g(s(s(s(s(s(s(s(s(T27)))))))))

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:

QA_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → U1_G(T27, pB_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
QA_IN_G(s(s(s(s(s(s(s(s(T27))))))))) → PB_IN_GG(T27, s(s(s(s(s(s(s(0))))))))
PB_IN_GG(s(T40), T41) → U2_GG(T40, T41, pB_in_gg(T40, s(T41)))
PB_IN_GG(s(T40), T41) → PB_IN_GG(T40, s(T41))

The TRS R consists of the following rules:

qA_in_g(0) → qA_out_g(0)
qA_in_g(s(0)) → qA_out_g(s(0))
qA_in_g(s(s(0))) → qA_out_g(s(s(0)))
qA_in_g(s(s(s(0)))) → qA_out_g(s(s(s(0))))
qA_in_g(s(s(s(s(0))))) → qA_out_g(s(s(s(s(0)))))
qA_in_g(s(s(s(s(s(0)))))) → qA_out_g(s(s(s(s(s(0))))))
qA_in_g(s(s(s(s(s(s(0))))))) → qA_out_g(s(s(s(s(s(s(0)))))))
qA_in_g(s(s(s(s(s(s(s(0)))))))) → qA_out_g(s(s(s(s(s(s(s(0))))))))
qA_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U1_g(T27, pB_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pB_in_gg(0, T35) → pB_out_gg(0, T35)
pB_in_gg(s(T40), T41) → U2_gg(T40, T41, pB_in_gg(T40, s(T41)))
U2_gg(T40, T41, pB_out_gg(T40, s(T41))) → pB_out_gg(s(T40), T41)
U1_g(T27, pB_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qA_out_g(s(s(s(s(s(s(s(s(T27)))))))))

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 3 less nodes.

(6) Obligation:

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

PB_IN_GG(s(T40), T41) → PB_IN_GG(T40, s(T41))

The TRS R consists of the following rules:

qA_in_g(0) → qA_out_g(0)
qA_in_g(s(0)) → qA_out_g(s(0))
qA_in_g(s(s(0))) → qA_out_g(s(s(0)))
qA_in_g(s(s(s(0)))) → qA_out_g(s(s(s(0))))
qA_in_g(s(s(s(s(0))))) → qA_out_g(s(s(s(s(0)))))
qA_in_g(s(s(s(s(s(0)))))) → qA_out_g(s(s(s(s(s(0))))))
qA_in_g(s(s(s(s(s(s(0))))))) → qA_out_g(s(s(s(s(s(s(0)))))))
qA_in_g(s(s(s(s(s(s(s(0)))))))) → qA_out_g(s(s(s(s(s(s(s(0))))))))
qA_in_g(s(s(s(s(s(s(s(s(T27))))))))) → U1_g(T27, pB_in_gg(T27, s(s(s(s(s(s(s(0)))))))))
pB_in_gg(0, T35) → pB_out_gg(0, T35)
pB_in_gg(s(T40), T41) → U2_gg(T40, T41, pB_in_gg(T40, s(T41)))
U2_gg(T40, T41, pB_out_gg(T40, s(T41))) → pB_out_gg(s(T40), T41)
U1_g(T27, pB_out_gg(T27, s(s(s(s(s(s(s(0))))))))) → qA_out_g(s(s(s(s(s(s(s(s(T27)))))))))

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:

PB_IN_GG(s(T40), T41) → PB_IN_GG(T40, s(T41))

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:

PB_IN_GG(s(T40), T41) → PB_IN_GG(T40, s(T41))

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

(12) YES