Left Termination of the query pattern p(f) w.r.t. the given Prolog program could not be shown:
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
p1(.2(A, {}0)) :- l1(.2(A, {}0)).
r1(10).
l1({}0).
l1(.2(H, T)) :- r1(H), l1(T).
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (f)
l1: (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_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_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_A1(._22(A, []_0)) -> IF_P_1_IN_1_A2(A, l_1_in_a1(._22(A, []_0)))
P_1_IN_A1(._22(A, []_0)) -> L_1_IN_A1(._22(A, []_0))
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
P_1_IN_A1(x1) = P_1_IN_A
IF_L_1_IN_2_A3(x1, x2, x3) = IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1) = R_1_IN_A
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_A1(._22(A, []_0)) -> IF_P_1_IN_1_A2(A, l_1_in_a1(._22(A, []_0)))
P_1_IN_A1(._22(A, []_0)) -> L_1_IN_A1(._22(A, []_0))
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
P_1_IN_A1(x1) = P_1_IN_A
IF_L_1_IN_2_A3(x1, x2, x3) = IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1) = R_1_IN_A
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.
↳ PROLOG
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
↳ PrologToPiTRSProof
Pi DP problem:
The TRS P consists of the following rules:
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
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:
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
The argument filtering Pi contains the following mapping:
._22(x1, x2) = ._22(x1, x2)
1_0 = 1_0
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
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:
L_1_IN_A -> IF_L_1_IN_1_A1(r_1_in_a)
IF_L_1_IN_1_A1(r_1_out_a1(H)) -> L_1_IN_A
The TRS R consists of the following rules:
r_1_in_a -> r_1_out_a1(1_0)
The set Q consists of the following terms:
r_1_in_a
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_L_1_IN_1_A1, L_1_IN_A}.
With regard to the inferred argument filtering the predicates were used in the following modes:
p1: (f)
l1: (f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_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_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
Pi DP problem:
The TRS P consists of the following rules:
P_1_IN_A1(._22(A, []_0)) -> IF_P_1_IN_1_A2(A, l_1_in_a1(._22(A, []_0)))
P_1_IN_A1(._22(A, []_0)) -> L_1_IN_A1(._22(A, []_0))
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
P_1_IN_A1(x1) = P_1_IN_A
IF_L_1_IN_2_A3(x1, x2, x3) = IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1) = R_1_IN_A
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_A1(._22(A, []_0)) -> IF_P_1_IN_1_A2(A, l_1_in_a1(._22(A, []_0)))
P_1_IN_A1(._22(A, []_0)) -> L_1_IN_A1(._22(A, []_0))
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
L_1_IN_A1(._22(H, T)) -> R_1_IN_A1(H)
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> IF_L_1_IN_2_A3(H, T, l_1_in_a1(T))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
P_1_IN_A1(x1) = P_1_IN_A
IF_L_1_IN_2_A3(x1, x2, x3) = IF_L_1_IN_2_A2(x1, x3)
IF_P_1_IN_1_A2(x1, x2) = IF_P_1_IN_1_A1(x2)
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
R_1_IN_A1(x1) = R_1_IN_A
We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 4 less nodes.
↳ PROLOG
↳ PrologToPiTRSProof
↳ PrologToPiTRSProof
↳ PiTRS
↳ DependencyPairsProof
↳ PiDP
↳ DependencyGraphProof
↳ PiDP
↳ UsableRulesProof
Pi DP problem:
The TRS P consists of the following rules:
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
p_1_in_a1(._22(A, []_0)) -> if_p_1_in_1_a2(A, l_1_in_a1(._22(A, []_0)))
l_1_in_a1([]_0) -> l_1_out_a1([]_0)
l_1_in_a1(._22(H, T)) -> if_l_1_in_1_a3(H, T, r_1_in_a1(H))
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
if_l_1_in_1_a3(H, T, r_1_out_a1(H)) -> if_l_1_in_2_a3(H, T, l_1_in_a1(T))
if_l_1_in_2_a3(H, T, l_1_out_a1(T)) -> l_1_out_a1(._22(H, T))
if_p_1_in_1_a2(A, l_1_out_a1(._22(A, []_0))) -> p_1_out_a1(._22(A, []_0))
The argument filtering Pi contains the following mapping:
p_1_in_a1(x1) = p_1_in_a
._22(x1, x2) = ._22(x1, x2)
[]_0 = []_0
1_0 = 1_0
if_p_1_in_1_a2(x1, x2) = if_p_1_in_1_a1(x2)
l_1_in_a1(x1) = l_1_in_a
l_1_out_a1(x1) = l_1_out_a1(x1)
if_l_1_in_1_a3(x1, x2, x3) = if_l_1_in_1_a1(x3)
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
if_l_1_in_2_a3(x1, x2, x3) = if_l_1_in_2_a2(x1, x3)
p_1_out_a1(x1) = p_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
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:
L_1_IN_A1(._22(H, T)) -> IF_L_1_IN_1_A3(H, T, r_1_in_a1(H))
IF_L_1_IN_1_A3(H, T, r_1_out_a1(H)) -> L_1_IN_A1(T)
The TRS R consists of the following rules:
r_1_in_a1(1_0) -> r_1_out_a1(1_0)
The argument filtering Pi contains the following mapping:
._22(x1, x2) = ._22(x1, x2)
1_0 = 1_0
r_1_in_a1(x1) = r_1_in_a
r_1_out_a1(x1) = r_1_out_a1(x1)
L_1_IN_A1(x1) = L_1_IN_A
IF_L_1_IN_1_A3(x1, x2, x3) = IF_L_1_IN_1_A1(x3)
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:
L_1_IN_A -> IF_L_1_IN_1_A1(r_1_in_a)
IF_L_1_IN_1_A1(r_1_out_a1(H)) -> L_1_IN_A
The TRS R consists of the following rules:
r_1_in_a -> r_1_out_a1(1_0)
The set Q consists of the following terms:
r_1_in_a
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_L_1_IN_1_A1, L_1_IN_A}.