(0) Obligation:
Clauses:
m(X, 0, Z) :- ','(!, =(Z, X)).
m(0, Y, Z) :- ','(!, =(Z, 0)).
m(X, Y, Z) :- ','(p(X, A), ','(p(Y, B), m(A, B, Z))).
p(0, 0).
p(s(X), X).
q(X, Y, Z) :- m(X, Y, Z).
q(X, Y, Z).
=(X, X).
Queries:
q(g,g,a).
(1) BuiltinConflictTransformerProof (SOUND transformation)
Renamed defined predicates conflicting with built-in predicates [PROLOG].
(2) Obligation:
Clauses:
m(X, 0, Z) :- ','(!, user_defined_=(Z, X)).
m(0, Y, Z) :- ','(!, user_defined_=(Z, 0)).
m(X, Y, Z) :- ','(p(X, A), ','(p(Y, B), m(A, B, Z))).
p(0, 0).
p(s(X), X).
q(X, Y, Z) :- m(X, Y, Z).
q(X, Y, Z).
user_defined_=(X, X).
Queries:
q(g,g,a).
(3) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(4) Obligation:
Triples:
m36(s(T109), s(T114), T105) :- m36(T109, T114, T105).
q1(s(T66), s(T71), T62) :- m36(T66, T71, T62).
Clauses:
mc36(T83, 0, T83).
mc36(0, T88, 0).
mc36(s(T109), s(T114), T105) :- mc36(T109, T114, T105).
Afs:
q1(x1, x2, x3) = q1(x1, x2)
(5) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
q1_in: (b,b,f)
m36_in: (b,b,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
Q1_IN_GGA(s(T66), s(T71), T62) → U2_GGA(T66, T71, T62, m36_in_gga(T66, T71, T62))
Q1_IN_GGA(s(T66), s(T71), T62) → M36_IN_GGA(T66, T71, T62)
M36_IN_GGA(s(T109), s(T114), T105) → U1_GGA(T109, T114, T105, m36_in_gga(T109, T114, T105))
M36_IN_GGA(s(T109), s(T114), T105) → M36_IN_GGA(T109, T114, T105)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
m36_in_gga(
x1,
x2,
x3) =
m36_in_gga(
x1,
x2)
Q1_IN_GGA(
x1,
x2,
x3) =
Q1_IN_GGA(
x1,
x2)
U2_GGA(
x1,
x2,
x3,
x4) =
U2_GGA(
x1,
x2,
x4)
M36_IN_GGA(
x1,
x2,
x3) =
M36_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
Q1_IN_GGA(s(T66), s(T71), T62) → U2_GGA(T66, T71, T62, m36_in_gga(T66, T71, T62))
Q1_IN_GGA(s(T66), s(T71), T62) → M36_IN_GGA(T66, T71, T62)
M36_IN_GGA(s(T109), s(T114), T105) → U1_GGA(T109, T114, T105, m36_in_gga(T109, T114, T105))
M36_IN_GGA(s(T109), s(T114), T105) → M36_IN_GGA(T109, T114, T105)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
m36_in_gga(
x1,
x2,
x3) =
m36_in_gga(
x1,
x2)
Q1_IN_GGA(
x1,
x2,
x3) =
Q1_IN_GGA(
x1,
x2)
U2_GGA(
x1,
x2,
x3,
x4) =
U2_GGA(
x1,
x2,
x4)
M36_IN_GGA(
x1,
x2,
x3) =
M36_IN_GGA(
x1,
x2)
U1_GGA(
x1,
x2,
x3,
x4) =
U1_GGA(
x1,
x2,
x4)
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:
M36_IN_GGA(s(T109), s(T114), T105) → M36_IN_GGA(T109, T114, T105)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
M36_IN_GGA(
x1,
x2,
x3) =
M36_IN_GGA(
x1,
x2)
We have to consider all (P,R,Pi)-chains
(9) PiDPToQDPProof (SOUND 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:
M36_IN_GGA(s(T109), s(T114)) → M36_IN_GGA(T109, T114)
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:
- M36_IN_GGA(s(T109), s(T114)) → M36_IN_GGA(T109, T114)
The graph contains the following edges 1 > 1, 2 > 2
(12) YES