Left Termination of the query pattern goal(b) w.r.t. the given Prolog program could not be shown:
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
p1(a0).
p1(X) :- p1(Y).
q1(b0).
goal1(X) :- p1(X), q1(X).
With regard to the inferred argument filtering the predicates were used in the following modes:
goal1: (b)
p1: (b) (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g1(x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g1(x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g
goal_1_out_g1(x1) = goal_1_out_g
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:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g1(x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g1(x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g
goal_1_out_g1(x1) = goal_1_out_g
Pi DP problem:
The TRS P consists of the following rules:
GOAL_1_IN_G1(X) -> IF_GOAL_1_IN_1_G2(X, p_1_in_g1(X))
GOAL_1_IN_G1(X) -> P_1_IN_G1(X)
P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_a1(Y))
P_1_IN_G1(X) -> P_1_IN_A1(Y)
P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, p_1_in_a1(Y))
P_1_IN_A1(X) -> P_1_IN_A1(Y)
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> IF_GOAL_1_IN_2_G2(X, q_1_in_g1(X))
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> Q_1_IN_G1(X)
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g1(x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g1(x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g
goal_1_out_g1(x1) = goal_1_out_g
GOAL_1_IN_G1(x1) = GOAL_1_IN_G1(x1)
Q_1_IN_G1(x1) = Q_1_IN_G1(x1)
P_1_IN_A1(x1) = P_1_IN_A
IF_GOAL_1_IN_1_G2(x1, x2) = IF_GOAL_1_IN_1_G2(x1, x2)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2) = IF_P_1_IN_1_G1(x2)
IF_GOAL_1_IN_2_G2(x1, x2) = IF_GOAL_1_IN_2_G1(x2)
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:
GOAL_1_IN_G1(X) -> IF_GOAL_1_IN_1_G2(X, p_1_in_g1(X))
GOAL_1_IN_G1(X) -> P_1_IN_G1(X)
P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_a1(Y))
P_1_IN_G1(X) -> P_1_IN_A1(Y)
P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, p_1_in_a1(Y))
P_1_IN_A1(X) -> P_1_IN_A1(Y)
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> IF_GOAL_1_IN_2_G2(X, q_1_in_g1(X))
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> Q_1_IN_G1(X)
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g1(x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g1(x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g
goal_1_out_g1(x1) = goal_1_out_g
GOAL_1_IN_G1(x1) = GOAL_1_IN_G1(x1)
Q_1_IN_G1(x1) = Q_1_IN_G1(x1)
P_1_IN_A1(x1) = P_1_IN_A
IF_GOAL_1_IN_1_G2(x1, x2) = IF_GOAL_1_IN_1_G2(x1, x2)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2) = IF_P_1_IN_1_G1(x2)
IF_GOAL_1_IN_2_G2(x1, x2) = IF_GOAL_1_IN_2_G1(x2)
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 7 less nodes.
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_A1(X) -> P_1_IN_A1(Y)
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g1(x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g1(x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g
goal_1_out_g1(x1) = goal_1_out_g
P_1_IN_A1(x1) = P_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:
P_1_IN_A1(X) -> P_1_IN_A1(Y)
R is empty.
The argument filtering Pi contains the following mapping:
P_1_IN_A1(x1) = P_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:
P_1_IN_A -> P_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 {P_1_IN_A}.
With regard to the inferred argument filtering the predicates were used in the following modes:
goal1: (b)
p1: (b) (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g2(x1, x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g2(x1, x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g1(x1)
goal_1_out_g1(x1) = goal_1_out_g1(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:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g2(x1, x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g2(x1, x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g1(x1)
goal_1_out_g1(x1) = goal_1_out_g1(x1)
Pi DP problem:
The TRS P consists of the following rules:
GOAL_1_IN_G1(X) -> IF_GOAL_1_IN_1_G2(X, p_1_in_g1(X))
GOAL_1_IN_G1(X) -> P_1_IN_G1(X)
P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_a1(Y))
P_1_IN_G1(X) -> P_1_IN_A1(Y)
P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, p_1_in_a1(Y))
P_1_IN_A1(X) -> P_1_IN_A1(Y)
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> IF_GOAL_1_IN_2_G2(X, q_1_in_g1(X))
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> Q_1_IN_G1(X)
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g2(x1, x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g2(x1, x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g1(x1)
goal_1_out_g1(x1) = goal_1_out_g1(x1)
GOAL_1_IN_G1(x1) = GOAL_1_IN_G1(x1)
Q_1_IN_G1(x1) = Q_1_IN_G1(x1)
P_1_IN_A1(x1) = P_1_IN_A
IF_GOAL_1_IN_1_G2(x1, x2) = IF_GOAL_1_IN_1_G2(x1, x2)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2) = IF_P_1_IN_1_G2(x1, x2)
IF_GOAL_1_IN_2_G2(x1, x2) = IF_GOAL_1_IN_2_G2(x1, x2)
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:
GOAL_1_IN_G1(X) -> IF_GOAL_1_IN_1_G2(X, p_1_in_g1(X))
GOAL_1_IN_G1(X) -> P_1_IN_G1(X)
P_1_IN_G1(X) -> IF_P_1_IN_1_G2(X, p_1_in_a1(Y))
P_1_IN_G1(X) -> P_1_IN_A1(Y)
P_1_IN_A1(X) -> IF_P_1_IN_1_A2(X, p_1_in_a1(Y))
P_1_IN_A1(X) -> P_1_IN_A1(Y)
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> IF_GOAL_1_IN_2_G2(X, q_1_in_g1(X))
IF_GOAL_1_IN_1_G2(X, p_1_out_g1(X)) -> Q_1_IN_G1(X)
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g2(x1, x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g2(x1, x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g1(x1)
goal_1_out_g1(x1) = goal_1_out_g1(x1)
GOAL_1_IN_G1(x1) = GOAL_1_IN_G1(x1)
Q_1_IN_G1(x1) = Q_1_IN_G1(x1)
P_1_IN_A1(x1) = P_1_IN_A
IF_GOAL_1_IN_1_G2(x1, x2) = IF_GOAL_1_IN_1_G2(x1, x2)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_P_1_IN_1_G2(x1, x2) = IF_P_1_IN_1_G2(x1, x2)
IF_GOAL_1_IN_2_G2(x1, x2) = IF_GOAL_1_IN_2_G2(x1, x2)
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 7 less nodes.
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_A1(X) -> P_1_IN_A1(Y)
The TRS R consists of the following rules:
goal_1_in_g1(X) -> if_goal_1_in_1_g2(X, p_1_in_g1(X))
p_1_in_g1(a_0) -> p_1_out_g1(a_0)
p_1_in_g1(X) -> if_p_1_in_1_g2(X, p_1_in_a1(Y))
p_1_in_a1(a_0) -> p_1_out_a1(a_0)
p_1_in_a1(X) -> if_p_1_in_1_a2(X, p_1_in_a1(Y))
if_p_1_in_1_a2(X, p_1_out_a1(Y)) -> p_1_out_a1(X)
if_p_1_in_1_g2(X, p_1_out_a1(Y)) -> p_1_out_g1(X)
if_goal_1_in_1_g2(X, p_1_out_g1(X)) -> if_goal_1_in_2_g2(X, q_1_in_g1(X))
q_1_in_g1(b_0) -> q_1_out_g1(b_0)
if_goal_1_in_2_g2(X, q_1_out_g1(X)) -> goal_1_out_g1(X)
The argument filtering Pi contains the following mapping:
goal_1_in_g1(x1) = goal_1_in_g1(x1)
a_0 = a_0
b_0 = b_0
if_goal_1_in_1_g2(x1, x2) = if_goal_1_in_1_g2(x1, x2)
p_1_in_g1(x1) = p_1_in_g1(x1)
p_1_out_g1(x1) = p_1_out_g1(x1)
if_p_1_in_1_g2(x1, x2) = if_p_1_in_1_g2(x1, x2)
p_1_in_a1(x1) = p_1_in_a
p_1_out_a1(x1) = p_1_out_a
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
if_goal_1_in_2_g2(x1, x2) = if_goal_1_in_2_g2(x1, x2)
q_1_in_g1(x1) = q_1_in_g1(x1)
q_1_out_g1(x1) = q_1_out_g1(x1)
goal_1_out_g1(x1) = goal_1_out_g1(x1)
P_1_IN_A1(x1) = P_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:
P_1_IN_A1(X) -> P_1_IN_A1(Y)
R is empty.
The argument filtering Pi contains the following mapping:
P_1_IN_A1(x1) = P_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:
P_1_IN_A -> P_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 {P_1_IN_A}.