Left Termination of the query pattern p2(f) w.r.t. the given Prolog program could not be shown:
↳ PROLOG
↳ UnrequestedClauseRemoverProof
p11(f1(X)) :- p11(X).
p21(f1(X)) :- p21(X).
The clause
p11(f1(X)) :- p11(X).
can be ignored, as it is not needed by any of the given querys.
Deleting this clauses results in the following prolog program:
p21(f1(X)) :- p21(X).
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
p21(f1(X)) :- p21(X).
With regard to the inferred argument filtering the predicates were used in the following modes:
p21: (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PrologToPiTRSProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> IF_P2_1_IN_1_A2(X, p2_1_in_a1(X))
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
IF_P2_1_IN_1_A2(x1, x2) = IF_P2_1_IN_1_A1(x2)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> IF_P2_1_IN_1_A2(X, p2_1_in_a1(X))
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
IF_P2_1_IN_1_A2(x1, x2) = IF_P2_1_IN_1_A1(x2)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
R is empty.
The argument filtering Pi contains the following mapping:
f_11(x1) = f_11(x1)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
P2_1_IN_A -> P2_1_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P2_1_IN_A}.
With regard to the inferred argument filtering the predicates were used in the following modes:
p21: (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> IF_P2_1_IN_1_A2(X, p2_1_in_a1(X))
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
IF_P2_1_IN_1_A2(x1, x2) = IF_P2_1_IN_1_A1(x2)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> IF_P2_1_IN_1_A2(X, p2_1_in_a1(X))
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
IF_P2_1_IN_1_A2(x1, x2) = IF_P2_1_IN_1_A1(x2)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
The TRS R consists of the following rules:
p2_1_in_a1(f_11(X)) -> if_p2_1_in_1_a2(X, p2_1_in_a1(X))
if_p2_1_in_1_a2(X, p2_1_out_a1(X)) -> p2_1_out_a1(f_11(X))
The argument filtering Pi contains the following mapping:
p2_1_in_a1(x1) = p2_1_in_a
f_11(x1) = f_11(x1)
if_p2_1_in_1_a2(x1, x2) = if_p2_1_in_1_a1(x2)
p2_1_out_a1(x1) = p2_1_out_a1(x1)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
P2_1_IN_A1(f_11(X)) -> P2_1_IN_A1(X)
R is empty.
The argument filtering Pi contains the following mapping:
f_11(x1) = f_11(x1)
P2_1_IN_A1(x1) = P2_1_IN_A
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.
↳ PROLOG
↳ UnrequestedClauseRemoverProof
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P2_1_IN_A -> P2_1_IN_A
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P2_1_IN_A}.