(0) Obligation:
Clauses:
max(X, Y, X) :- less(Y, X).
max(X, Y, Y) :- less(X, s(Y)).
less(0, s(X1)).
less(s(X), s(Y)) :- less(X, Y).
Query: max(a,a,g)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
lessA(s(X1), s(X2)) :- lessA(X1, X2).
maxB(s(X1), s(X2), s(X1)) :- lessA(X2, X1).
maxB(X1, X2, X2) :- lessA(X1, s(X2)).
maxB(s(X1), X2, X2) :- lessA(X1, X2).
Clauses:
lesscA(0, s(X1)).
lesscA(s(X1), s(X2)) :- lesscA(X1, X2).
Afs:
maxB(x1, x2, x3) = maxB(x3)
(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:
maxB_in: (f,f,b)
lessA_in: (f,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
MAXB_IN_AAG(s(X1), s(X2), s(X1)) → U2_AAG(X1, X2, lessA_in_ag(X2, X1))
MAXB_IN_AAG(s(X1), s(X2), s(X1)) → LESSA_IN_AG(X2, X1)
LESSA_IN_AG(s(X1), s(X2)) → U1_AG(X1, X2, lessA_in_ag(X1, X2))
LESSA_IN_AG(s(X1), s(X2)) → LESSA_IN_AG(X1, X2)
MAXB_IN_AAG(X1, X2, X2) → U3_AAG(X1, X2, lessA_in_ag(X1, s(X2)))
MAXB_IN_AAG(X1, X2, X2) → LESSA_IN_AG(X1, s(X2))
MAXB_IN_AAG(s(X1), X2, X2) → U4_AAG(X1, X2, lessA_in_ag(X1, X2))
MAXB_IN_AAG(s(X1), X2, X2) → LESSA_IN_AG(X1, X2)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
lessA_in_ag(
x1,
x2) =
lessA_in_ag(
x2)
MAXB_IN_AAG(
x1,
x2,
x3) =
MAXB_IN_AAG(
x3)
U2_AAG(
x1,
x2,
x3) =
U2_AAG(
x1,
x3)
LESSA_IN_AG(
x1,
x2) =
LESSA_IN_AG(
x2)
U1_AG(
x1,
x2,
x3) =
U1_AG(
x2,
x3)
U3_AAG(
x1,
x2,
x3) =
U3_AAG(
x2,
x3)
U4_AAG(
x1,
x2,
x3) =
U4_AAG(
x2,
x3)
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:
MAXB_IN_AAG(s(X1), s(X2), s(X1)) → U2_AAG(X1, X2, lessA_in_ag(X2, X1))
MAXB_IN_AAG(s(X1), s(X2), s(X1)) → LESSA_IN_AG(X2, X1)
LESSA_IN_AG(s(X1), s(X2)) → U1_AG(X1, X2, lessA_in_ag(X1, X2))
LESSA_IN_AG(s(X1), s(X2)) → LESSA_IN_AG(X1, X2)
MAXB_IN_AAG(X1, X2, X2) → U3_AAG(X1, X2, lessA_in_ag(X1, s(X2)))
MAXB_IN_AAG(X1, X2, X2) → LESSA_IN_AG(X1, s(X2))
MAXB_IN_AAG(s(X1), X2, X2) → U4_AAG(X1, X2, lessA_in_ag(X1, X2))
MAXB_IN_AAG(s(X1), X2, X2) → LESSA_IN_AG(X1, X2)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
lessA_in_ag(
x1,
x2) =
lessA_in_ag(
x2)
MAXB_IN_AAG(
x1,
x2,
x3) =
MAXB_IN_AAG(
x3)
U2_AAG(
x1,
x2,
x3) =
U2_AAG(
x1,
x3)
LESSA_IN_AG(
x1,
x2) =
LESSA_IN_AG(
x2)
U1_AG(
x1,
x2,
x3) =
U1_AG(
x2,
x3)
U3_AAG(
x1,
x2,
x3) =
U3_AAG(
x2,
x3)
U4_AAG(
x1,
x2,
x3) =
U4_AAG(
x2,
x3)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 7 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
LESSA_IN_AG(s(X1), s(X2)) → LESSA_IN_AG(X1, X2)
R is empty.
The argument filtering Pi contains the following mapping:
s(
x1) =
s(
x1)
LESSA_IN_AG(
x1,
x2) =
LESSA_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LESSA_IN_AG(s(X2)) → LESSA_IN_AG(X2)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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:
- LESSA_IN_AG(s(X2)) → LESSA_IN_AG(X2)
The graph contains the following edges 1 > 1
(10) YES