Left Termination of the query pattern p(b) w.r.t. the given Prolog program could not be shown:
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
p1(b0).
p1(a0) :- p11(X).
p11(b0).
p11(a0) :- p11(X).
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (b)
p11: (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PrologToPiTRSProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_G1(a_0) -> IF_P_1_IN_1_G1(p1_1_in_a1(X))
P_1_IN_G1(a_0) -> P1_1_IN_A1(X)
P1_1_IN_A1(a_0) -> IF_P1_1_IN_1_A1(p1_1_in_a1(X))
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
IF_P1_1_IN_1_A1(x1) = IF_P1_1_IN_1_A1(x1)
IF_P_1_IN_1_G1(x1) = IF_P_1_IN_1_G1(x1)
P1_1_IN_A1(x1) = P1_1_IN_A
P_1_IN_G1(x1) = P_1_IN_G1(x1)
We have to consider all (P,R,Pi)-chains
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_G1(a_0) -> IF_P_1_IN_1_G1(p1_1_in_a1(X))
P_1_IN_G1(a_0) -> P1_1_IN_A1(X)
P1_1_IN_A1(a_0) -> IF_P1_1_IN_1_A1(p1_1_in_a1(X))
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
IF_P1_1_IN_1_A1(x1) = IF_P1_1_IN_1_A1(x1)
IF_P_1_IN_1_G1(x1) = IF_P_1_IN_1_G1(x1)
P1_1_IN_A1(x1) = P1_1_IN_A
P_1_IN_G1(x1) = P_1_IN_G1(x1)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
P1_1_IN_A1(x1) = P1_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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
R is empty.
The argument filtering Pi contains the following mapping:
a_0 = a_0
P1_1_IN_A1(x1) = P1_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
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
↳ PrologToPiTRSProof
Q DP problem:
The TRS P consists of the following rules:
P1_1_IN_A -> P1_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 {P1_1_IN_A}.
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (b)
p11: (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_G1(a_0) -> IF_P_1_IN_1_G1(p1_1_in_a1(X))
P_1_IN_G1(a_0) -> P1_1_IN_A1(X)
P1_1_IN_A1(a_0) -> IF_P1_1_IN_1_A1(p1_1_in_a1(X))
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
IF_P1_1_IN_1_A1(x1) = IF_P1_1_IN_1_A1(x1)
IF_P_1_IN_1_G1(x1) = IF_P_1_IN_1_G1(x1)
P1_1_IN_A1(x1) = P1_1_IN_A
P_1_IN_G1(x1) = P_1_IN_G1(x1)
We have to consider all (P,R,Pi)-chains
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_G1(a_0) -> IF_P_1_IN_1_G1(p1_1_in_a1(X))
P_1_IN_G1(a_0) -> P1_1_IN_A1(X)
P1_1_IN_A1(a_0) -> IF_P1_1_IN_1_A1(p1_1_in_a1(X))
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
IF_P1_1_IN_1_A1(x1) = IF_P1_1_IN_1_A1(x1)
IF_P_1_IN_1_G1(x1) = IF_P_1_IN_1_G1(x1)
P1_1_IN_A1(x1) = P1_1_IN_A
P_1_IN_G1(x1) = P_1_IN_G1(x1)
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
The TRS R consists of the following rules:
p_1_in_g1(b_0) -> p_1_out_g1(b_0)
p_1_in_g1(a_0) -> if_p_1_in_1_g1(p1_1_in_a1(X))
p1_1_in_a1(b_0) -> p1_1_out_a1(b_0)
p1_1_in_a1(a_0) -> if_p1_1_in_1_a1(p1_1_in_a1(X))
if_p1_1_in_1_a1(p1_1_out_a1(X)) -> p1_1_out_a1(a_0)
if_p_1_in_1_g1(p1_1_out_a1(X)) -> p_1_out_g1(a_0)
The argument filtering Pi contains the following mapping:
p_1_in_g1(x1) = p_1_in_g1(x1)
b_0 = b_0
a_0 = a_0
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g1(x1) = if_p_1_in_1_g1(x1)
p1_1_in_a1(x1) = p1_1_in_a
p1_1_out_a1(x1) = p1_1_out_a1(x1)
if_p1_1_in_1_a1(x1) = if_p1_1_in_1_a1(x1)
P1_1_IN_A1(x1) = P1_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
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
Pi DP problem:
The TRS P consists of the following rules:
P1_1_IN_A1(a_0) -> P1_1_IN_A1(X)
R is empty.
The argument filtering Pi contains the following mapping:
a_0 = a_0
P1_1_IN_A1(x1) = P1_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
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PiDP
↳ PiDPToQDPProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P1_1_IN_A -> P1_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 {P1_1_IN_A}.