Left Termination proof of /tmp/tmpJOD4RB/p

Left Termination of the query pattern p(b) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

p₁(cons₂(X, nil₀)).
p₁(cons₂(s₁²(X), cons₂(Y, Xs))) :- p₁(cons₂(X, cons₂(Y, Xs))), mult₃(X, Y, Z), p₁(cons₂(Z, Xs)).
p₁(cons₂(0₀, Xs)) :- p₁(Xs).
sum₃(X, 0₀, X).
sum₃(X, s₁(Y), s₁(Z)) :- sum₃(X, Y, Z).
mult₃(underscore, 0₀, 0₀).
mult₃(X, s₁(Y), Z) :- mult₃(X, Y, W), sum₃(W, X, Z).

With regard to the inferred argument filtering the predicates were used in the following modes:
p₁: (b)
mult₃: (b,b,f)
sum₃: (b,b,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(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> IF_P_1_IN_1_G₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))
P_1_IN_G₁(cons_2₂(0_0, Xs)) -> IF_P_1_IN_4_G₂(Xs, p_1_in_g₁(Xs))
P_1_IN_G₁(cons_2₂(0_0, Xs)) -> P_1_IN_G₁(Xs)
IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> MULT_3_IN_GGA₃(X, Y, Z)
MULT_3_IN_GGA₃(X, s_1₁(Y), Z) -> IF_MULT_3_IN_1_GGA₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
MULT_3_IN_GGA₃(X, s_1₁(Y), Z) -> MULT_3_IN_GGA₃(X, Y, W)
IF_MULT_3_IN_1_GGA₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> IF_MULT_3_IN_2_GGA₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
IF_MULT_3_IN_1_GGA₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> SUM_3_IN_GGA₃(W, X, Z)
SUM_3_IN_GGA₃(X, s_1₁(Y), s_1₁(Z)) -> IF_SUM_3_IN_1_GGA₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
SUM_3_IN_GGA₃(X, s_1₁(Y), s_1₁(Z)) -> SUM_3_IN_GGA₃(X, Y, Z)
IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> IF_P_1_IN_3_G₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> P_1_IN_G₁(cons_2₂(Z, Xs))

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g₁(x1) = p_1_in_g₁(x1)
cons_2₂(x1, x2) = cons_2₂(x1, x2)
nil_0 = nil_0
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
p_1_out_g₁(x1) = p_1_out_g
if_p_1_in_1_g₄(x1, x2, x3, x4) = if_p_1_in_1_g₄(x1, x2, x3, x4)
if_p_1_in_4_g₂(x1, x2) = if_p_1_in_4_g₁(x2)
if_p_1_in_2_g₄(x1, x2, x3, x4) = if_p_1_in_2_g₂(x3, x4)
mult_3_in_gga₃(x1, x2, x3) = mult_3_in_gga₂(x1, x2)
mult_3_out_gga₃(x1, x2, x3) = mult_3_out_gga₁(x3)
if_mult_3_in_1_gga₄(x1, x2, x3, x4) = if_mult_3_in_1_gga₂(x1, x4)
if_mult_3_in_2_gga₅(x1, x2, x3, x4, x5) = if_mult_3_in_2_gga₁(x5)
sum_3_in_gga₃(x1, x2, x3) = sum_3_in_gga₂(x1, x2)
sum_3_out_gga₃(x1, x2, x3) = sum_3_out_gga₁(x3)
if_sum_3_in_1_gga₄(x1, x2, x3, x4) = if_sum_3_in_1_gga₁(x4)
if_p_1_in_3_g₅(x1, x2, x3, x4, x5) = if_p_1_in_3_g₁(x5)
IF_SUM_3_IN_1_GGA₄(x1, x2, x3, x4) = IF_SUM_3_IN_1_GGA₁(x4)
IF_MULT_3_IN_2_GGA₅(x1, x2, x3, x4, x5) = IF_MULT_3_IN_2_GGA₁(x5)
MULT_3_IN_GGA₃(x1, x2, x3) = MULT_3_IN_GGA₂(x1, x2)
IF_P_1_IN_2_G₄(x1, x2, x3, x4) = IF_P_1_IN_2_G₂(x3, x4)
SUM_3_IN_GGA₃(x1, x2, x3) = SUM_3_IN_GGA₂(x1, x2)
IF_MULT_3_IN_1_GGA₄(x1, x2, x3, x4) = IF_MULT_3_IN_1_GGA₂(x1, x4)
IF_P_1_IN_4_G₂(x1, x2) = IF_P_1_IN_4_G₁(x2)
IF_P_1_IN_3_G₅(x1, x2, x3, x4, x5) = IF_P_1_IN_3_G₁(x5)
IF_P_1_IN_1_G₄(x1, x2, x3, x4) = IF_P_1_IN_1_G₄(x1, x2, x3, x4)
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

Pi DP problem:
The TRS P consists of the following rules:

P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> IF_P_1_IN_1_G₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))
P_1_IN_G₁(cons_2₂(0_0, Xs)) -> IF_P_1_IN_4_G₂(Xs, p_1_in_g₁(Xs))
P_1_IN_G₁(cons_2₂(0_0, Xs)) -> P_1_IN_G₁(Xs)
IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> MULT_3_IN_GGA₃(X, Y, Z)
MULT_3_IN_GGA₃(X, s_1₁(Y), Z) -> IF_MULT_3_IN_1_GGA₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
MULT_3_IN_GGA₃(X, s_1₁(Y), Z) -> MULT_3_IN_GGA₃(X, Y, W)
IF_MULT_3_IN_1_GGA₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> IF_MULT_3_IN_2_GGA₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
IF_MULT_3_IN_1_GGA₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> SUM_3_IN_GGA₃(W, X, Z)
SUM_3_IN_GGA₃(X, s_1₁(Y), s_1₁(Z)) -> IF_SUM_3_IN_1_GGA₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
SUM_3_IN_GGA₃(X, s_1₁(Y), s_1₁(Z)) -> SUM_3_IN_GGA₃(X, Y, Z)
IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> IF_P_1_IN_3_G₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> P_1_IN_G₁(cons_2₂(Z, Xs))

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

SUM_3_IN_GGA₃(X, s_1₁(Y), s_1₁(Z)) -> SUM_3_IN_GGA₃(X, Y, Z)

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

SUM_3_IN_GGA₃(X, s_1₁(Y), s_1₁(Z)) -> SUM_3_IN_GGA₃(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_1₁(x1) = s_1₁(x1)
SUM_3_IN_GGA₃(x1, x2, x3) = SUM_3_IN_GGA₂(x1, x2)

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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

SUM_3_IN_GGA₂(X, s_1₁(Y)) -> SUM_3_IN_GGA₂(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {SUM_3_IN_GGA₂}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

SUM_3_IN_GGA₂(X, s_1₁(Y)) -> SUM_3_IN_GGA₂(X, Y)
The graph contains the following edges 1 >= 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

MULT_3_IN_GGA₃(X, s_1₁(Y), Z) -> MULT_3_IN_GGA₃(X, Y, W)

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

MULT_3_IN_GGA₃(X, s_1₁(Y), Z) -> MULT_3_IN_GGA₃(X, Y, W)

R is empty.
The argument filtering Pi contains the following mapping:
s_1₁(x1) = s_1₁(x1)
MULT_3_IN_GGA₃(x1, x2, x3) = MULT_3_IN_GGA₂(x1, x2)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

MULT_3_IN_GGA₂(X, s_1₁(Y)) -> MULT_3_IN_GGA₂(X, Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {MULT_3_IN_GGA₂}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

MULT_3_IN_GGA₂(X, s_1₁(Y)) -> MULT_3_IN_GGA₂(X, Y)
The graph contains the following edges 1 >= 1, 2 > 2

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
P_1_IN_G₁(cons_2₂(0_0, Xs)) -> P_1_IN_G₁(Xs)
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> IF_P_1_IN_1_G₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
IF_P_1_IN_2_G₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> P_1_IN_G₁(cons_2₂(Z, Xs))
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g₁(cons_2₂(X, nil_0))
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₂(Xs, p_1_in_g₁(Xs))
if_p_1_in_4_g₂(Xs, p_1_out_g₁(Xs)) -> p_1_out_g₁(cons_2₂(0_0, Xs))
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g₁(cons_2₂(X, cons_2₂(Y, Xs)))) -> if_p_1_in_2_g₄(X, Y, Xs, mult_3_in_gga₃(X, Y, Z))
mult_3_in_gga₃(underscore, 0_0, 0_0) -> mult_3_out_gga₃(underscore, 0_0, 0_0)
mult_3_in_gga₃(X, s_1₁(Y), Z) -> if_mult_3_in_1_gga₄(X, Y, Z, mult_3_in_gga₃(X, Y, W))
if_mult_3_in_1_gga₄(X, Y, Z, mult_3_out_gga₃(X, Y, W)) -> if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_in_gga₃(W, X, Z))
sum_3_in_gga₃(X, 0_0, X) -> sum_3_out_gga₃(X, 0_0, X)
sum_3_in_gga₃(X, s_1₁(Y), s_1₁(Z)) -> if_sum_3_in_1_gga₄(X, Y, Z, sum_3_in_gga₃(X, Y, Z))
if_sum_3_in_1_gga₄(X, Y, Z, sum_3_out_gga₃(X, Y, Z)) -> sum_3_out_gga₃(X, s_1₁(Y), s_1₁(Z))
if_mult_3_in_2_gga₅(X, Y, Z, W, sum_3_out_gga₃(W, X, Z)) -> mult_3_out_gga₃(X, s_1₁(Y), Z)
if_p_1_in_2_g₄(X, Y, Xs, mult_3_out_gga₃(X, Y, Z)) -> if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₅(X, Y, Xs, Z, p_1_out_g₁(cons_2₂(Z, Xs))) -> p_1_out_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs)))

The argument filtering Pi contains the following mapping:
p_1_in_g₁(x1) = p_1_in_g₁(x1)
cons_2₂(x1, x2) = cons_2₂(x1, x2)
nil_0 = nil_0
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
p_1_out_g₁(x1) = p_1_out_g
if_p_1_in_1_g₄(x1, x2, x3, x4) = if_p_1_in_1_g₄(x1, x2, x3, x4)
if_p_1_in_4_g₂(x1, x2) = if_p_1_in_4_g₁(x2)
if_p_1_in_2_g₄(x1, x2, x3, x4) = if_p_1_in_2_g₂(x3, x4)
mult_3_in_gga₃(x1, x2, x3) = mult_3_in_gga₂(x1, x2)
mult_3_out_gga₃(x1, x2, x3) = mult_3_out_gga₁(x3)
if_mult_3_in_1_gga₄(x1, x2, x3, x4) = if_mult_3_in_1_gga₂(x1, x4)
if_mult_3_in_2_gga₅(x1, x2, x3, x4, x5) = if_mult_3_in_2_gga₁(x5)
sum_3_in_gga₃(x1, x2, x3) = sum_3_in_gga₂(x1, x2)
sum_3_out_gga₃(x1, x2, x3) = sum_3_out_gga₁(x3)
if_sum_3_in_1_gga₄(x1, x2, x3, x4) = if_sum_3_in_1_gga₁(x4)
if_p_1_in_3_g₅(x1, x2, x3, x4, x5) = if_p_1_in_3_g₁(x5)
IF_P_1_IN_2_G₄(x1, x2, x3, x4) = IF_P_1_IN_2_G₂(x3, x4)
IF_P_1_IN_1_G₄(x1, x2, x3, x4) = IF_P_1_IN_1_G₄(x1, x2, x3, x4)
P_1_IN_G₁(x1) = P_1_IN_G₁(x1)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

Q DP problem:
The TRS P consists of the following rules:

IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g) -> IF_P_1_IN_2_G₂(Xs, mult_3_in_gga₂(X, Y))
P_1_IN_G₁(cons_2₂(0_0, Xs)) -> P_1_IN_G₁(Xs)
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> IF_P_1_IN_1_G₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
IF_P_1_IN_2_G₂(Xs, mult_3_out_gga₁(Z)) -> P_1_IN_G₁(cons_2₂(Z, Xs))
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₁(p_1_in_g₁(Xs))
if_p_1_in_4_g₁(p_1_out_g) -> p_1_out_g
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g) -> if_p_1_in_2_g₂(Xs, mult_3_in_gga₂(X, Y))
mult_3_in_gga₂(underscore, 0_0) -> mult_3_out_gga₁(0_0)
mult_3_in_gga₂(X, s_1₁(Y)) -> if_mult_3_in_1_gga₂(X, mult_3_in_gga₂(X, Y))
if_mult_3_in_1_gga₂(X, mult_3_out_gga₁(W)) -> if_mult_3_in_2_gga₁(sum_3_in_gga₂(W, X))
sum_3_in_gga₂(X, 0_0) -> sum_3_out_gga₁(X)
sum_3_in_gga₂(X, s_1₁(Y)) -> if_sum_3_in_1_gga₁(sum_3_in_gga₂(X, Y))
if_sum_3_in_1_gga₁(sum_3_out_gga₁(Z)) -> sum_3_out_gga₁(s_1₁(Z))
if_mult_3_in_2_gga₁(sum_3_out_gga₁(Z)) -> mult_3_out_gga₁(Z)
if_p_1_in_2_g₂(Xs, mult_3_out_gga₁(Z)) -> if_p_1_in_3_g₁(p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₁(p_1_out_g) -> p_1_out_g

The set Q consists of the following terms:

p_1_in_g₁(x0)
if_p_1_in_4_g₁(x0)
if_p_1_in_1_g₄(x0, x1, x2, x3)
mult_3_in_gga₂(x0, x1)
if_mult_3_in_1_gga₂(x0, x1)
sum_3_in_gga₂(x0, x1)
if_sum_3_in_1_gga₁(x0)
if_mult_3_in_2_gga₁(x0)
if_p_1_in_2_g₂(x0, x1)
if_p_1_in_3_g₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_P_1_IN_2_G₂, IF_P_1_IN_1_G₄, P_1_IN_G₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

P_1_IN_G₁(cons_2₂(0_0, Xs)) -> P_1_IN_G₁(Xs)
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> IF_P_1_IN_1_G₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))

The remaining Dependency Pairs were at least non-strictly be oriented.

IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g) -> IF_P_1_IN_2_G₂(Xs, mult_3_in_gga₂(X, Y))
IF_P_1_IN_2_G₂(Xs, mult_3_out_gga₁(Z)) -> P_1_IN_G₁(cons_2₂(Z, Xs))
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))

With the implicit AFS there is no usable rule.

Used ordering: POLO with Polynomial interpretation:

POL(if_mult_3_in_2_gga₁(x₁)) = 0
POL(0_0) = 0
POL(IF_P_1_IN_1_G₄(x₁, x₂, x₃, x₄)) = 1 + x₃
POL(if_p_1_in_2_g₂(x₁, x₂)) = 0
POL(sum_3_out_gga₁(x₁)) = 0
POL(mult_3_in_gga₂(x₁, x₂)) = 0
POL(if_p_1_in_3_g₁(x₁)) = 0
POL(p_1_in_g₁(x₁)) = 0
POL(P_1_IN_G₁(x₁)) = x₁
POL(p_1_out_g) = 0
POL(if_mult_3_in_1_gga₂(x₁, x₂)) = 0
POL(nil_0) = 0
POL(if_p_1_in_4_g₁(x₁)) = 0
POL(if_sum_3_in_1_gga₁(x₁)) = 0
POL(IF_P_1_IN_2_G₂(x₁, x₂)) = 1 + x₁
POL(sum_3_in_gga₂(x₁, x₂)) = 0
POL(s_1₁(x₁)) = 0
POL(if_p_1_in_1_g₄(x₁, x₂, x₃, x₄)) = 0
POL(cons_2₂(x₁, x₂)) = 1 + x₂
POL(mult_3_out_gga₁(x₁)) = 0

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

IF_P_1_IN_1_G₄(X, Y, Xs, p_1_out_g) -> IF_P_1_IN_2_G₂(Xs, mult_3_in_gga₂(X, Y))
IF_P_1_IN_2_G₂(Xs, mult_3_out_gga₁(Z)) -> P_1_IN_G₁(cons_2₂(Z, Xs))
P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₁(p_1_in_g₁(Xs))
if_p_1_in_4_g₁(p_1_out_g) -> p_1_out_g
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g) -> if_p_1_in_2_g₂(Xs, mult_3_in_gga₂(X, Y))
mult_3_in_gga₂(underscore, 0_0) -> mult_3_out_gga₁(0_0)
mult_3_in_gga₂(X, s_1₁(Y)) -> if_mult_3_in_1_gga₂(X, mult_3_in_gga₂(X, Y))
if_mult_3_in_1_gga₂(X, mult_3_out_gga₁(W)) -> if_mult_3_in_2_gga₁(sum_3_in_gga₂(W, X))
sum_3_in_gga₂(X, 0_0) -> sum_3_out_gga₁(X)
sum_3_in_gga₂(X, s_1₁(Y)) -> if_sum_3_in_1_gga₁(sum_3_in_gga₂(X, Y))
if_sum_3_in_1_gga₁(sum_3_out_gga₁(Z)) -> sum_3_out_gga₁(s_1₁(Z))
if_mult_3_in_2_gga₁(sum_3_out_gga₁(Z)) -> mult_3_out_gga₁(Z)
if_p_1_in_2_g₂(Xs, mult_3_out_gga₁(Z)) -> if_p_1_in_3_g₁(p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₁(p_1_out_g) -> p_1_out_g

The set Q consists of the following terms:

p_1_in_g₁(x0)
if_p_1_in_4_g₁(x0)
if_p_1_in_1_g₄(x0, x1, x2, x3)
mult_3_in_gga₂(x0, x1)
if_mult_3_in_1_gga₂(x0, x1)
sum_3_in_gga₂(x0, x1)
if_sum_3_in_1_gga₁(x0)
if_mult_3_in_2_gga₁(x0)
if_p_1_in_2_g₂(x0, x1)
if_p_1_in_3_g₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_P_1_IN_2_G₂, IF_P_1_IN_1_G₄, P_1_IN_G₁}.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))

The TRS R consists of the following rules:

p_1_in_g₁(cons_2₂(X, nil_0)) -> p_1_out_g
p_1_in_g₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> if_p_1_in_1_g₄(X, Y, Xs, p_1_in_g₁(cons_2₂(X, cons_2₂(Y, Xs))))
p_1_in_g₁(cons_2₂(0_0, Xs)) -> if_p_1_in_4_g₁(p_1_in_g₁(Xs))
if_p_1_in_4_g₁(p_1_out_g) -> p_1_out_g
if_p_1_in_1_g₄(X, Y, Xs, p_1_out_g) -> if_p_1_in_2_g₂(Xs, mult_3_in_gga₂(X, Y))
mult_3_in_gga₂(underscore, 0_0) -> mult_3_out_gga₁(0_0)
mult_3_in_gga₂(X, s_1₁(Y)) -> if_mult_3_in_1_gga₂(X, mult_3_in_gga₂(X, Y))
if_mult_3_in_1_gga₂(X, mult_3_out_gga₁(W)) -> if_mult_3_in_2_gga₁(sum_3_in_gga₂(W, X))
sum_3_in_gga₂(X, 0_0) -> sum_3_out_gga₁(X)
sum_3_in_gga₂(X, s_1₁(Y)) -> if_sum_3_in_1_gga₁(sum_3_in_gga₂(X, Y))
if_sum_3_in_1_gga₁(sum_3_out_gga₁(Z)) -> sum_3_out_gga₁(s_1₁(Z))
if_mult_3_in_2_gga₁(sum_3_out_gga₁(Z)) -> mult_3_out_gga₁(Z)
if_p_1_in_2_g₂(Xs, mult_3_out_gga₁(Z)) -> if_p_1_in_3_g₁(p_1_in_g₁(cons_2₂(Z, Xs)))
if_p_1_in_3_g₁(p_1_out_g) -> p_1_out_g

The set Q consists of the following terms:

p_1_in_g₁(x0)
if_p_1_in_4_g₁(x0)
if_p_1_in_1_g₄(x0, x1, x2, x3)
mult_3_in_gga₂(x0, x1)
if_mult_3_in_1_gga₂(x0, x1)
sum_3_in_gga₂(x0, x1)
if_sum_3_in_1_gga₁(x0)
if_mult_3_in_2_gga₁(x0)
if_p_1_in_2_g₂(x0, x1)
if_p_1_in_3_g₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G₁}.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, we can delete all non-usable rules from R.

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPPoloProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs)))

R is empty.
The set Q consists of the following terms:

p_1_in_g₁(x0)
if_p_1_in_4_g₁(x0)
if_p_1_in_1_g₄(x0, x1, x2, x3)
mult_3_in_gga₂(x0, x1)
if_mult_3_in_1_gga₂(x0, x1)
sum_3_in_gga₂(x0, x1)
if_sum_3_in_1_gga₁(x0)
if_mult_3_in_2_gga₁(x0)
if_p_1_in_2_g₂(x0, x1)
if_p_1_in_3_g₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {P_1_IN_G₁}.
We used the following order and afs together with the size-change analysis to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s_1₁(x1) = s_1₁(x1)
cons_2₂(x1, x2) = x1

From the DPs we obtained the following set of size-change graphs:

P_1_IN_G₁(cons_2₂(s_1₁(s_1₁(X)), cons_2₂(Y, Xs))) -> P_1_IN_G₁(cons_2₂(X, cons_2₂(Y, Xs))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1

We oriented the following set of usable rules. none