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

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇒, 86 ms)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇒, 37 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 PiDP
        ↳8 PiDPToQDPProof (⇒, 0 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 PiDP
        ↳13 PiDPToQDPProof (⇒, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES

(0) Obligation:

Clauses:

q(X, Y) :- m(X, Y, Z).
m(X, 0, X).
m(0, Y, 0) :- !.
m(X, Y, Z) :- ','(p(X, A), ','(p(Y, B), m(A, B, Z))).
p(0, 0).
p(s(X), X).

Query: q(g,g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

mA(s(X1), X2) :- mA(X1, X2).
mB(s(X1), 0, X2) :- mA(X1, X2).
mB(s(X1), s(X2), X3) :- mB(X1, X2, X3).
qC(X1, X2) :- mB(X1, X2, X3).

Clauses:

mcA(X1, X1).
mcA(0, 0).
mcA(s(X1), X2) :- mcA(X1, X2).
mcB(X1, 0, X1).
mcB(0, 0, 0).
mcB(s(X1), 0, X2) :- mcA(X1, X2).
mcB(0, X1, 0).
mcB(s(X1), s(X2), X3) :- mcB(X1, X2, X3).

Afs:

qC(x1, x2)  =  qC(x1, x2)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
qC_in: (b,b)
mB_in: (b,b,f)
mA_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

QC_IN_GG(X1, X2) → U4_GG(X1, X2, mB_in_gga(X1, X2, X3))
QC_IN_GG(X1, X2) → MB_IN_GGA(X1, X2, X3)
MB_IN_GGA(s(X1), 0, X2) → U2_GGA(X1, X2, mA_in_ga(X1, X2))
MB_IN_GGA(s(X1), 0, X2) → MA_IN_GA(X1, X2)
MA_IN_GA(s(X1), X2) → U1_GA(X1, X2, mA_in_ga(X1, X2))
MA_IN_GA(s(X1), X2) → MA_IN_GA(X1, X2)
MB_IN_GGA(s(X1), s(X2), X3) → U3_GGA(X1, X2, X3, mB_in_gga(X1, X2, X3))
MB_IN_GGA(s(X1), s(X2), X3) → MB_IN_GGA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
s(x1)  =  s(x1)
0  =  0
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
QC_IN_GG(x1, x2)  =  QC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
MB_IN_GGA(x1, x2, x3)  =  MB_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3)  =  U2_GGA(x1, x3)
MA_IN_GA(x1, x2)  =  MA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x1, x2, x4)

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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

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

QC_IN_GG(X1, X2) → U4_GG(X1, X2, mB_in_gga(X1, X2, X3))
QC_IN_GG(X1, X2) → MB_IN_GGA(X1, X2, X3)
MB_IN_GGA(s(X1), 0, X2) → U2_GGA(X1, X2, mA_in_ga(X1, X2))
MB_IN_GGA(s(X1), 0, X2) → MA_IN_GA(X1, X2)
MA_IN_GA(s(X1), X2) → U1_GA(X1, X2, mA_in_ga(X1, X2))
MA_IN_GA(s(X1), X2) → MA_IN_GA(X1, X2)
MB_IN_GGA(s(X1), s(X2), X3) → U3_GGA(X1, X2, X3, mB_in_gga(X1, X2, X3))
MB_IN_GGA(s(X1), s(X2), X3) → MB_IN_GGA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
mB_in_gga(x1, x2, x3)  =  mB_in_gga(x1, x2)
s(x1)  =  s(x1)
0  =  0
mA_in_ga(x1, x2)  =  mA_in_ga(x1)
QC_IN_GG(x1, x2)  =  QC_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
MB_IN_GGA(x1, x2, x3)  =  MB_IN_GGA(x1, x2)
U2_GGA(x1, x2, x3)  =  U2_GGA(x1, x3)
MA_IN_GA(x1, x2)  =  MA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U3_GGA(x1, x2, x3, x4)  =  U3_GGA(x1, x2, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MA_IN_GA(s(X1), X2) → MA_IN_GA(X1, X2)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
MA_IN_GA(x1, x2)  =  MA_IN_GA(x1)

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

(8) PiDPToQDPProof (SOUND transformation)

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

(9) Obligation:

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

MA_IN_GA(s(X1)) → MA_IN_GA(X1)

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

(10) 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:

  • MA_IN_GA(s(X1)) → MA_IN_GA(X1)
    The graph contains the following edges 1 > 1

(11) YES

(12) Obligation:

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

MB_IN_GGA(s(X1), s(X2), X3) → MB_IN_GGA(X1, X2, X3)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
MB_IN_GGA(x1, x2, x3)  =  MB_IN_GGA(x1, x2)

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

(13) PiDPToQDPProof (SOUND transformation)

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

(14) Obligation:

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

MB_IN_GGA(s(X1), s(X2)) → MB_IN_GGA(X1, X2)

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

(15) 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:

  • MB_IN_GGA(s(X1), s(X2)) → MB_IN_GGA(X1, X2)
    The graph contains the following edges 1 > 1, 2 > 2

(16) YES