Left Termination proof of /tmp/tmpYgehV3/perm.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

ap1₃(nil₀, X, X).
ap1₃(cons₂(H, X), Y, cons₂(H, Z)) :- ap1₃(X, Y, Z).
ap2₃(nil₀, X, X).
ap2₃(cons₂(H, X), Y, cons₂(H, Z)) :- ap2₃(X, Y, Z).
perm₂(nil₀, nil₀).
perm₂(Xs, cons₂(X, Ys)) :- ap1₃(X1s, cons₂(X, X2s), Xs), ap2₃(X1s, X2s, Zs), perm₂(Zs, Ys).

With regard to the inferred argument filtering the predicates were used in the following modes:
perm₂: (b,f)
ap1₃: (f,f,b)
ap2₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

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:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

PERM_2_IN_GA₂(Xs, cons_2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
PERM_2_IN_GA₂(Xs, cons_2₂(X, Ys)) -> AP1_3_IN_AAG₃(X1s, cons_2₂(X, X2s), Xs)
AP1_3_IN_AAG₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> IF_AP1_3_IN_1_AAG₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
AP1_3_IN_AAG₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP1_3_IN_AAG₃(X, Y, Z)
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> AP2_3_IN_GGA₃(X1s, X2s, Zs)
AP2_3_IN_GGA₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> IF_AP2_3_IN_1_GGA₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
AP2_3_IN_GGA₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP2_3_IN_GGA₃(X, Y, Z)
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_GA₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₁(x4)
ap1_3_in_aag₃(x1, x2, x3) = ap1_3_in_aag₁(x3)
ap1_3_out_aag₃(x1, x2, x3) = ap1_3_out_aag₂(x1, x2)
if_ap1_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_ap1_3_in_1_aag₂(x1, x5)
if_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x2, x6)
ap2_3_in_gga₃(x1, x2, x3) = ap2_3_in_gga₂(x1, x2)
ap2_3_out_gga₃(x1, x2, x3) = ap2_3_out_gga₁(x3)
if_ap2_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_ap2_3_in_1_gga₂(x1, x5)
if_perm_2_in_3_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ga₂(x2, x5)
AP2_3_IN_GGA₃(x1, x2, x3) = AP2_3_IN_GGA₂(x1, x2)
IF_PERM_2_IN_3_GA₅(x1, x2, x3, x4, x5) = IF_PERM_2_IN_3_GA₂(x2, x5)
AP1_3_IN_AAG₃(x1, x2, x3) = AP1_3_IN_AAG₁(x3)
IF_AP2_3_IN_1_GGA₅(x1, x2, x3, x4, x5) = IF_AP2_3_IN_1_GGA₂(x1, x5)
IF_AP1_3_IN_1_AAG₅(x1, x2, x3, x4, x5) = IF_AP1_3_IN_1_AAG₂(x1, x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x2, x6)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₁(x4)

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:

PERM_2_IN_GA₂(Xs, cons_2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
PERM_2_IN_GA₂(Xs, cons_2₂(X, Ys)) -> AP1_3_IN_AAG₃(X1s, cons_2₂(X, X2s), Xs)
AP1_3_IN_AAG₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> IF_AP1_3_IN_1_AAG₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
AP1_3_IN_AAG₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP1_3_IN_AAG₃(X, Y, Z)
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> AP2_3_IN_GGA₃(X1s, X2s, Zs)
AP2_3_IN_GGA₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> IF_AP2_3_IN_1_GGA₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
AP2_3_IN_GGA₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP2_3_IN_GGA₃(X, Y, Z)
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> IF_PERM_2_IN_3_GA₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

AP2_3_IN_GGA₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP2_3_IN_GGA₃(X, Y, Z)

The TRS R consists of the following rules:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

AP2_3_IN_GGA₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP2_3_IN_GGA₃(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons_2₂(x1, x2) = cons_2₂(x1, x2)
AP2_3_IN_GGA₃(x1, x2, x3) = AP2_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:

AP2_3_IN_GGA₂(cons_2₂(H, X), Y) -> AP2_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 {AP2_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:

AP2_3_IN_GGA₂(cons_2₂(H, X), Y) -> AP2_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:

AP1_3_IN_AAG₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP1_3_IN_AAG₃(X, Y, Z)

The TRS R consists of the following rules:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

AP1_3_IN_AAG₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> AP1_3_IN_AAG₃(X, Y, Z)

R is empty.
The argument filtering Pi contains the following mapping:
cons_2₂(x1, x2) = cons_2₂(x1, x2)
AP1_3_IN_AAG₃(x1, x2, x3) = AP1_3_IN_AAG₁(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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

AP1_3_IN_AAG₁(cons_2₂(H, Z)) -> AP1_3_IN_AAG₁(Z)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {AP1_3_IN_AAG₁}.
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:

AP1_3_IN_AAG₁(cons_2₂(H, Z)) -> AP1_3_IN_AAG₁(Z)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
PERM_2_IN_GA₂(Xs, cons_2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)

The TRS R consists of the following rules:

perm_2_in_ga₂(nil_0, nil_0) -> perm_2_out_ga₂(nil_0, nil_0)
perm_2_in_ga₂(Xs, cons_2₂(X, Ys)) -> if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_1_ga₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_perm_2_in_2_ga₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_in_ga₂(Zs, Ys))
if_perm_2_in_3_ga₅(Xs, X, Ys, Zs, perm_2_out_ga₂(Zs, Ys)) -> perm_2_out_ga₂(Xs, cons_2₂(X, Ys))

The argument filtering Pi contains the following mapping:
perm_2_in_ga₂(x1, x2) = perm_2_in_ga₁(x1)
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
perm_2_out_ga₂(x1, x2) = perm_2_out_ga₁(x2)
if_perm_2_in_1_ga₄(x1, x2, x3, x4) = if_perm_2_in_1_ga₁(x4)
ap1_3_in_aag₃(x1, x2, x3) = ap1_3_in_aag₁(x3)
ap1_3_out_aag₃(x1, x2, x3) = ap1_3_out_aag₂(x1, x2)
if_ap1_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_ap1_3_in_1_aag₂(x1, x5)
if_perm_2_in_2_ga₆(x1, x2, x3, x4, x5, x6) = if_perm_2_in_2_ga₂(x2, x6)
ap2_3_in_gga₃(x1, x2, x3) = ap2_3_in_gga₂(x1, x2)
ap2_3_out_gga₃(x1, x2, x3) = ap2_3_out_gga₁(x3)
if_ap2_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_ap2_3_in_1_gga₂(x1, x5)
if_perm_2_in_3_ga₅(x1, x2, x3, x4, x5) = if_perm_2_in_3_ga₂(x2, x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x2, x6)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₁(x4)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_out_aag₃(X1s, cons_2₂(X, X2s), Xs)) -> IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_in_gga₃(X1s, X2s, Zs))
PERM_2_IN_GA₂(Xs, cons_2₂(X, Ys)) -> IF_PERM_2_IN_1_GA₄(Xs, X, Ys, ap1_3_in_aag₃(X1s, cons_2₂(X, X2s), Xs))
IF_PERM_2_IN_2_GA₆(Xs, X, Ys, X1s, X2s, ap2_3_out_gga₃(X1s, X2s, Zs)) -> PERM_2_IN_GA₂(Zs, Ys)

The TRS R consists of the following rules:

ap2_3_in_gga₃(nil_0, X, X) -> ap2_3_out_gga₃(nil_0, X, X)
ap2_3_in_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_in_gga₃(X, Y, Z))
ap1_3_in_aag₃(nil_0, X, X) -> ap1_3_out_aag₃(nil_0, X, X)
ap1_3_in_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_in_aag₃(X, Y, Z))
if_ap2_3_in_1_gga₅(H, X, Y, Z, ap2_3_out_gga₃(X, Y, Z)) -> ap2_3_out_gga₃(cons_2₂(H, X), Y, cons_2₂(H, Z))
if_ap1_3_in_1_aag₅(H, X, Y, Z, ap1_3_out_aag₃(X, Y, Z)) -> ap1_3_out_aag₃(cons_2₂(H, X), Y, cons_2₂(H, Z))

The argument filtering Pi contains the following mapping:
nil_0 = nil_0
cons_2₂(x1, x2) = cons_2₂(x1, x2)
ap1_3_in_aag₃(x1, x2, x3) = ap1_3_in_aag₁(x3)
ap1_3_out_aag₃(x1, x2, x3) = ap1_3_out_aag₂(x1, x2)
if_ap1_3_in_1_aag₅(x1, x2, x3, x4, x5) = if_ap1_3_in_1_aag₂(x1, x5)
ap2_3_in_gga₃(x1, x2, x3) = ap2_3_in_gga₂(x1, x2)
ap2_3_out_gga₃(x1, x2, x3) = ap2_3_out_gga₁(x3)
if_ap2_3_in_1_gga₅(x1, x2, x3, x4, x5) = if_ap2_3_in_1_gga₂(x1, x5)
PERM_2_IN_GA₂(x1, x2) = PERM_2_IN_GA₁(x1)
IF_PERM_2_IN_2_GA₆(x1, x2, x3, x4, x5, x6) = IF_PERM_2_IN_2_GA₂(x2, x6)
IF_PERM_2_IN_1_GA₄(x1, x2, x3, x4) = IF_PERM_2_IN_1_GA₁(x4)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

IF_PERM_2_IN_1_GA₁(ap1_3_out_aag₂(X1s, cons_2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, ap2_3_in_gga₂(X1s, X2s))
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(ap1_3_in_aag₁(Xs))
IF_PERM_2_IN_2_GA₂(X, ap2_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)

The TRS R consists of the following rules:

ap2_3_in_gga₂(nil_0, X) -> ap2_3_out_gga₁(X)
ap2_3_in_gga₂(cons_2₂(H, X), Y) -> if_ap2_3_in_1_gga₂(H, ap2_3_in_gga₂(X, Y))
ap1_3_in_aag₁(X) -> ap1_3_out_aag₂(nil_0, X)
ap1_3_in_aag₁(cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₂(H, ap1_3_in_aag₁(Z))
if_ap2_3_in_1_gga₂(H, ap2_3_out_gga₁(Z)) -> ap2_3_out_gga₁(cons_2₂(H, Z))
if_ap1_3_in_1_aag₂(H, ap1_3_out_aag₂(X, Y)) -> ap1_3_out_aag₂(cons_2₂(H, X), Y)

The set Q consists of the following terms:

ap2_3_in_gga₂(x0, x1)
ap1_3_in_aag₁(x0)
if_ap2_3_in_1_gga₂(x0, x1)
if_ap1_3_in_1_aag₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_2_GA₂, IF_PERM_2_IN_1_GA₁, PERM_2_IN_GA₁}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

ap2_3_in_gga₂(nil_0, X) -> ap2_3_out_gga₁(X)

Used ordering: POLO with Polynomial interpretation:

POL(nil_0) = 2
POL(ap2_3_out_gga₁(x₁)) = 1 + x₁
POL(if_ap1_3_in_1_aag₂(x₁, x₂)) = 1 + x₁ + x₂
POL(IF_PERM_2_IN_2_GA₂(x₁, x₂)) = 1 + x₁ + x₂
POL(ap2_3_in_gga₂(x₁, x₂)) = x₁ + x₂
POL(IF_PERM_2_IN_1_GA₁(x₁)) = x₁
POL(ap1_3_in_aag₁(x₁)) = 2 + x₁
POL(if_ap2_3_in_1_gga₂(x₁, x₂)) = 1 + x₁ + x₂
POL(cons_2₂(x₁, x₂)) = 1 + x₁ + x₂
POL(ap1_3_out_aag₂(x₁, x₂)) = x₁ + x₂
POL(PERM_2_IN_GA₁(x₁)) = 2 + x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPPoloProof

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

IF_PERM_2_IN_1_GA₁(ap1_3_out_aag₂(X1s, cons_2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, ap2_3_in_gga₂(X1s, X2s))
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(ap1_3_in_aag₁(Xs))
IF_PERM_2_IN_2_GA₂(X, ap2_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)

The TRS R consists of the following rules:

ap2_3_in_gga₂(cons_2₂(H, X), Y) -> if_ap2_3_in_1_gga₂(H, ap2_3_in_gga₂(X, Y))
ap1_3_in_aag₁(X) -> ap1_3_out_aag₂(nil_0, X)
ap1_3_in_aag₁(cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₂(H, ap1_3_in_aag₁(Z))
if_ap2_3_in_1_gga₂(H, ap2_3_out_gga₁(Z)) -> ap2_3_out_gga₁(cons_2₂(H, Z))
if_ap1_3_in_1_aag₂(H, ap1_3_out_aag₂(X, Y)) -> ap1_3_out_aag₂(cons_2₂(H, X), Y)

The set Q consists of the following terms:

ap2_3_in_gga₂(x0, x1)
ap1_3_in_aag₁(x0)
if_ap2_3_in_1_gga₂(x0, x1)
if_ap1_3_in_1_aag₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_2_GA₂, IF_PERM_2_IN_1_GA₁, PERM_2_IN_GA₁}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

IF_PERM_2_IN_2_GA₂(X, ap2_3_out_gga₁(Zs)) -> PERM_2_IN_GA₁(Zs)

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

IF_PERM_2_IN_1_GA₁(ap1_3_out_aag₂(X1s, cons_2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, ap2_3_in_gga₂(X1s, X2s))
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(ap1_3_in_aag₁(Xs))

With the implicit AFS we had to orient the following set of usable rules non-strictly.

if_ap2_3_in_1_gga₂(H, ap2_3_out_gga₁(Z)) -> ap2_3_out_gga₁(cons_2₂(H, Z))
ap2_3_in_gga₂(cons_2₂(H, X), Y) -> if_ap2_3_in_1_gga₂(H, ap2_3_in_gga₂(X, Y))

Used ordering: POLO with Polynomial interpretation:

POL(nil_0) = 0
POL(ap2_3_out_gga₁(x₁)) = 1
POL(if_ap1_3_in_1_aag₂(x₁, x₂)) = 0
POL(IF_PERM_2_IN_2_GA₂(x₁, x₂)) = x₂
POL(ap2_3_in_gga₂(x₁, x₂)) = 0
POL(IF_PERM_2_IN_1_GA₁(x₁)) = 0
POL(ap1_3_in_aag₁(x₁)) = 0
POL(if_ap2_3_in_1_gga₂(x₁, x₂)) = x₂
POL(cons_2₂(x₁, x₂)) = 0
POL(ap1_3_out_aag₂(x₁, x₂)) = 0
POL(PERM_2_IN_GA₁(x₁)) = 0

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ QDPPoloProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

IF_PERM_2_IN_1_GA₁(ap1_3_out_aag₂(X1s, cons_2₂(X, X2s))) -> IF_PERM_2_IN_2_GA₂(X, ap2_3_in_gga₂(X1s, X2s))
PERM_2_IN_GA₁(Xs) -> IF_PERM_2_IN_1_GA₁(ap1_3_in_aag₁(Xs))

The TRS R consists of the following rules:

ap2_3_in_gga₂(cons_2₂(H, X), Y) -> if_ap2_3_in_1_gga₂(H, ap2_3_in_gga₂(X, Y))
ap1_3_in_aag₁(X) -> ap1_3_out_aag₂(nil_0, X)
ap1_3_in_aag₁(cons_2₂(H, Z)) -> if_ap1_3_in_1_aag₂(H, ap1_3_in_aag₁(Z))
if_ap2_3_in_1_gga₂(H, ap2_3_out_gga₁(Z)) -> ap2_3_out_gga₁(cons_2₂(H, Z))
if_ap1_3_in_1_aag₂(H, ap1_3_out_aag₂(X, Y)) -> ap1_3_out_aag₂(cons_2₂(H, X), Y)

The set Q consists of the following terms:

ap2_3_in_gga₂(x0, x1)
ap1_3_in_aag₁(x0)
if_ap2_3_in_1_gga₂(x0, x1)
if_ap1_3_in_1_aag₂(x0, x1)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PERM_2_IN_2_GA₂, IF_PERM_2_IN_1_GA₁, PERM_2_IN_GA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.