(0) Obligation:
Clauses:
f(X) :- ','(p(X), q(X)).
p(a) :- !.
p(X) :- p(Y).
q(b).
Queries:
f(a).
(1) CutEliminatorProof (SOUND transformation)
Eliminated all cuts by simply ignoring them[PROLOG].
(2) Obligation:
Clauses:
f(X) :- ','(p(X), q(X)).
p(a) :- true.
p(X) :- p(Y).
q(b).
Queries:
f(a).
(3) UndefinedPredicateHandlerProof (SOUND transformation)
Added facts for all undefined predicates [PROLOG].
(4) Obligation:
Clauses:
f(X) :- ','(p(X), q(X)).
p(a) :- true.
p(X) :- p(Y).
q(b).
true.
Queries:
f(a).
(5) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
f_in: (f)
p_in: (f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_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:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
(7) 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:
F_IN_A(X) → U1_A(X, p_in_a(X))
F_IN_A(X) → P_IN_A(X)
P_IN_A(a) → U3_A(true_in_)
P_IN_A(a) → TRUE_IN_
P_IN_A(X) → U4_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U1_A(X, p_out_a(X)) → U2_A(X, q_in_a(X))
U1_A(X, p_out_a(X)) → Q_IN_A(X)
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
F_IN_A(
x1) =
F_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
P_IN_A(
x1) =
P_IN_A
U3_A(
x1) =
U3_A(
x1)
TRUE_IN_ =
TRUE_IN_
U4_A(
x1,
x2) =
U4_A(
x2)
U2_A(
x1,
x2) =
U2_A(
x2)
Q_IN_A(
x1) =
Q_IN_A
We have to consider all (P,R,Pi)-chains
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
F_IN_A(X) → U1_A(X, p_in_a(X))
F_IN_A(X) → P_IN_A(X)
P_IN_A(a) → U3_A(true_in_)
P_IN_A(a) → TRUE_IN_
P_IN_A(X) → U4_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U1_A(X, p_out_a(X)) → U2_A(X, q_in_a(X))
U1_A(X, p_out_a(X)) → Q_IN_A(X)
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
F_IN_A(
x1) =
F_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
P_IN_A(
x1) =
P_IN_A
U3_A(
x1) =
U3_A(
x1)
TRUE_IN_ =
TRUE_IN_
U4_A(
x1,
x2) =
U4_A(
x2)
U2_A(
x1,
x2) =
U2_A(
x2)
Q_IN_A(
x1) =
Q_IN_A
We have to consider all (P,R,Pi)-chains
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 7 less nodes.
(10) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_A(X) → P_IN_A(Y)
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
P_IN_A(
x1) =
P_IN_A
We have to consider all (P,R,Pi)-chains
(11) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(12) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_A(X) → P_IN_A(Y)
R is empty.
The argument filtering Pi contains the following mapping:
P_IN_A(
x1) =
P_IN_A
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:
P_IN_A → P_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(15) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
P_IN_A evaluates to t =
P_IN_AThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from P_IN_A to P_IN_A.
(16) FALSE
(17) PrologToPiTRSProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
f_in: (f)
p_in: (f)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(18) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
(19) 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:
F_IN_A(X) → U1_A(X, p_in_a(X))
F_IN_A(X) → P_IN_A(X)
P_IN_A(a) → U3_A(true_in_)
P_IN_A(a) → TRUE_IN_
P_IN_A(X) → U4_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U1_A(X, p_out_a(X)) → U2_A(X, q_in_a(X))
U1_A(X, p_out_a(X)) → Q_IN_A(X)
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
F_IN_A(
x1) =
F_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
P_IN_A(
x1) =
P_IN_A
U3_A(
x1) =
U3_A(
x1)
TRUE_IN_ =
TRUE_IN_
U4_A(
x1,
x2) =
U4_A(
x2)
U2_A(
x1,
x2) =
U2_A(
x2)
Q_IN_A(
x1) =
Q_IN_A
We have to consider all (P,R,Pi)-chains
(20) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
F_IN_A(X) → U1_A(X, p_in_a(X))
F_IN_A(X) → P_IN_A(X)
P_IN_A(a) → U3_A(true_in_)
P_IN_A(a) → TRUE_IN_
P_IN_A(X) → U4_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U1_A(X, p_out_a(X)) → U2_A(X, q_in_a(X))
U1_A(X, p_out_a(X)) → Q_IN_A(X)
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
F_IN_A(
x1) =
F_IN_A
U1_A(
x1,
x2) =
U1_A(
x2)
P_IN_A(
x1) =
P_IN_A
U3_A(
x1) =
U3_A(
x1)
TRUE_IN_ =
TRUE_IN_
U4_A(
x1,
x2) =
U4_A(
x2)
U2_A(
x1,
x2) =
U2_A(
x2)
Q_IN_A(
x1) =
Q_IN_A
We have to consider all (P,R,Pi)-chains
(21) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 7 less nodes.
(22) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_A(X) → P_IN_A(Y)
The TRS R consists of the following rules:
f_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(a) → U3_a(true_in_)
true_in_ → true_out_
U3_a(true_out_) → p_out_a(a)
p_in_a(X) → U4_a(X, p_in_a(Y))
U4_a(X, p_out_a(Y)) → p_out_a(X)
U1_a(X, p_out_a(X)) → U2_a(X, q_in_a(X))
q_in_a(b) → q_out_a(b)
U2_a(X, q_out_a(X)) → f_out_a(X)
The argument filtering Pi contains the following mapping:
f_in_a(
x1) =
f_in_a
U1_a(
x1,
x2) =
U1_a(
x2)
p_in_a(
x1) =
p_in_a
U3_a(
x1) =
U3_a(
x1)
true_in_ =
true_in_
true_out_ =
true_out_
p_out_a(
x1) =
p_out_a
U4_a(
x1,
x2) =
U4_a(
x2)
U2_a(
x1,
x2) =
U2_a(
x2)
q_in_a(
x1) =
q_in_a
q_out_a(
x1) =
q_out_a(
x1)
f_out_a(
x1) =
f_out_a(
x1)
P_IN_A(
x1) =
P_IN_A
We have to consider all (P,R,Pi)-chains
(23) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(24) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
P_IN_A(X) → P_IN_A(Y)
R is empty.
The argument filtering Pi contains the following mapping:
P_IN_A(
x1) =
P_IN_A
We have to consider all (P,R,Pi)-chains
(25) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P_IN_A → P_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(27) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
P_IN_A evaluates to t =
P_IN_AThus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from P_IN_A to P_IN_A.
(28) FALSE