(0) Obligation:
Clauses:
p(X) :- ','(q(X), r(X)).
p(X) :- =(X, 2).
q(X) :- ','(=(X, 1), !).
q(X) :- =(X, 3).
r(X) :- ','(=(X, 1), !).
r(X) :- =(X, 4).
=(X, X).
Query: p(a)
(1) BuiltinConflictTransformerProof (EQUIVALENT transformation)
Renamed defined predicates conflicting with built-in predicates [PROLOG].
(2) Obligation:
Clauses:
p(X) :- ','(q(X), r(X)).
p(X) :- user_defined_=(X, 2).
q(X) :- ','(user_defined_=(X, 1), !).
q(X) :- user_defined_=(X, 3).
r(X) :- ','(user_defined_=(X, 1), !).
r(X) :- user_defined_=(X, 4).
user_defined_=(X, X).
Query: p(a)
(3) PrologToPrologProblemTransformerProof (SOUND transformation)
Built Prolog problem from termination graph ICLP10.
(4) Obligation:
Clauses:
pA(1).
pA(2).
pA(2).
Query: pA(a)
(5) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
pA_in_a(1) → pA_out_a(1)
pA_in_a(2) → pA_out_a(2)
The argument filtering Pi contains the following mapping:
pA_in_a(
x1) =
pA_in_a
pA_out_a(
x1) =
pA_out_a(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(6) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
pA_in_a(1) → pA_out_a(1)
pA_in_a(2) → pA_out_a(2)
The argument filtering Pi contains the following mapping:
pA_in_a(
x1) =
pA_in_a
pA_out_a(
x1) =
pA_out_a(
x1)
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
P is empty.
The TRS R consists of the following rules:
pA_in_a(1) → pA_out_a(1)
pA_in_a(2) → pA_out_a(2)
The argument filtering Pi contains the following mapping:
pA_in_a(
x1) =
pA_in_a
pA_out_a(
x1) =
pA_out_a(
x1)
We have to consider all (P,R,Pi)-chains
(8) Obligation:
Pi DP problem:
P is empty.
The TRS R consists of the following rules:
pA_in_a(1) → pA_out_a(1)
pA_in_a(2) → pA_out_a(2)
The argument filtering Pi contains the following mapping:
pA_in_a(
x1) =
pA_in_a
pA_out_a(
x1) =
pA_out_a(
x1)
We have to consider all (P,R,Pi)-chains
(9) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,R,Pi) chain.
(10) YES