Left Termination proof of /tmp/tmpF-D9mm/intlist.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

intlist₂({}₀, {}₀).
intlist₂(.₂(X, XS), .₂(s₁(X), YS)) :- intlist₂(XS, YS).
int₃(0₀, 0₀, .₂(0₀, {}₀)).
int₃(0₀, s₁(Y), .₂(0₀, XS)) :- int₃(s₁(0₀), s₁(Y), XS).
int₃(s₁(X), 0₀, {}₀).
int₃(s₁(X), s₁(Y), XS) :- int₃(X, Y, ZS), intlist₂(ZS, XS).

With regard to the inferred argument filtering the predicates were used in the following modes:
int₃: (b,b,f)
intlist₂: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

int_3_in_gga₃(0_0, 0_0, ._2₂(0_0, []_0)) -> int_3_out_gga₃(0_0, 0_0, ._2₂(0_0, []_0))
int_3_in_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> if_int_3_in_1_gga₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
int_3_in_gga₃(s_1₁(X), 0_0, []_0) -> int_3_out_gga₃(s_1₁(X), 0_0, []_0)
int_3_in_gga₃(s_1₁(X), s_1₁(Y), XS) -> if_int_3_in_2_gga₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
if_int_3_in_2_gga₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
intlist_2_in_ga₂([]_0, []_0) -> intlist_2_out_ga₂([]_0, []_0)
intlist_2_in_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_out_ga₂(XS, YS)) -> intlist_2_out_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS))
if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_out_ga₂(ZS, XS)) -> int_3_out_gga₃(s_1₁(X), s_1₁(Y), XS)
if_int_3_in_1_gga₃(Y, XS, int_3_out_gga₃(s_1₁(0_0), s_1₁(Y), XS)) -> int_3_out_gga₃(0_0, s_1₁(Y), ._2₂(0_0, 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:

int_3_in_gga₃(0_0, 0_0, ._2₂(0_0, []_0)) -> int_3_out_gga₃(0_0, 0_0, ._2₂(0_0, []_0))
int_3_in_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> if_int_3_in_1_gga₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
int_3_in_gga₃(s_1₁(X), 0_0, []_0) -> int_3_out_gga₃(s_1₁(X), 0_0, []_0)
int_3_in_gga₃(s_1₁(X), s_1₁(Y), XS) -> if_int_3_in_2_gga₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
if_int_3_in_2_gga₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
intlist_2_in_ga₂([]_0, []_0) -> intlist_2_out_ga₂([]_0, []_0)
intlist_2_in_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_out_ga₂(XS, YS)) -> intlist_2_out_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS))
if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_out_ga₂(ZS, XS)) -> int_3_out_gga₃(s_1₁(X), s_1₁(Y), XS)
if_int_3_in_1_gga₃(Y, XS, int_3_out_gga₃(s_1₁(0_0), s_1₁(Y), XS)) -> int_3_out_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS))

INT_3_IN_GGA₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> IF_INT_3_IN_1_GGA₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
INT_3_IN_GGA₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> INT_3_IN_GGA₃(s_1₁(0_0), s_1₁(Y), XS)
INT_3_IN_GGA₃(s_1₁(X), s_1₁(Y), XS) -> IF_INT_3_IN_2_GGA₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
INT_3_IN_GGA₃(s_1₁(X), s_1₁(Y), XS) -> INT_3_IN_GGA₃(X, Y, ZS)
IF_INT_3_IN_2_GGA₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> IF_INT_3_IN_3_GGA₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
IF_INT_3_IN_2_GGA₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> INTLIST_2_IN_GA₂(ZS, XS)
INTLIST_2_IN_GA₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> IF_INTLIST_2_IN_1_GA₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
INTLIST_2_IN_GA₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> INTLIST_2_IN_GA₂(XS, YS)

The TRS R consists of the following rules:

int_3_in_gga₃(0_0, 0_0, ._2₂(0_0, []_0)) -> int_3_out_gga₃(0_0, 0_0, ._2₂(0_0, []_0))
int_3_in_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> if_int_3_in_1_gga₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
int_3_in_gga₃(s_1₁(X), 0_0, []_0) -> int_3_out_gga₃(s_1₁(X), 0_0, []_0)
int_3_in_gga₃(s_1₁(X), s_1₁(Y), XS) -> if_int_3_in_2_gga₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
if_int_3_in_2_gga₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
intlist_2_in_ga₂([]_0, []_0) -> intlist_2_out_ga₂([]_0, []_0)
intlist_2_in_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_out_ga₂(XS, YS)) -> intlist_2_out_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS))
if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_out_ga₂(ZS, XS)) -> int_3_out_gga₃(s_1₁(X), s_1₁(Y), XS)
if_int_3_in_1_gga₃(Y, XS, int_3_out_gga₃(s_1₁(0_0), s_1₁(Y), XS)) -> int_3_out_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS))

The argument filtering Pi contains the following mapping:
int_3_in_gga₃(x1, x2, x3) = int_3_in_gga₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
int_3_out_gga₃(x1, x2, x3) = int_3_out_gga₁(x3)
if_int_3_in_1_gga₃(x1, x2, x3) = if_int_3_in_1_gga₁(x3)
if_int_3_in_2_gga₄(x1, x2, x3, x4) = if_int_3_in_2_gga₁(x4)
if_int_3_in_3_gga₅(x1, x2, x3, x4, x5) = if_int_3_in_3_gga₁(x5)
intlist_2_in_ga₂(x1, x2) = intlist_2_in_ga₁(x1)
intlist_2_out_ga₂(x1, x2) = intlist_2_out_ga₁(x2)
if_intlist_2_in_1_ga₄(x1, x2, x3, x4) = if_intlist_2_in_1_ga₂(x1, x4)
INT_3_IN_GGA₃(x1, x2, x3) = INT_3_IN_GGA₂(x1, x2)
IF_INT_3_IN_3_GGA₅(x1, x2, x3, x4, x5) = IF_INT_3_IN_3_GGA₁(x5)
INTLIST_2_IN_GA₂(x1, x2) = INTLIST_2_IN_GA₁(x1)
IF_INT_3_IN_2_GGA₄(x1, x2, x3, x4) = IF_INT_3_IN_2_GGA₁(x4)
IF_INT_3_IN_1_GGA₃(x1, x2, x3) = IF_INT_3_IN_1_GGA₁(x3)
IF_INTLIST_2_IN_1_GA₄(x1, x2, x3, x4) = IF_INTLIST_2_IN_1_GA₂(x1, 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:

INT_3_IN_GGA₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> IF_INT_3_IN_1_GGA₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
INT_3_IN_GGA₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> INT_3_IN_GGA₃(s_1₁(0_0), s_1₁(Y), XS)
INT_3_IN_GGA₃(s_1₁(X), s_1₁(Y), XS) -> IF_INT_3_IN_2_GGA₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
INT_3_IN_GGA₃(s_1₁(X), s_1₁(Y), XS) -> INT_3_IN_GGA₃(X, Y, ZS)
IF_INT_3_IN_2_GGA₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> IF_INT_3_IN_3_GGA₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
IF_INT_3_IN_2_GGA₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> INTLIST_2_IN_GA₂(ZS, XS)
INTLIST_2_IN_GA₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> IF_INTLIST_2_IN_1_GA₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
INTLIST_2_IN_GA₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> INTLIST_2_IN_GA₂(XS, YS)

The TRS R consists of the following rules:

int_3_in_gga₃(0_0, 0_0, ._2₂(0_0, []_0)) -> int_3_out_gga₃(0_0, 0_0, ._2₂(0_0, []_0))
int_3_in_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> if_int_3_in_1_gga₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
int_3_in_gga₃(s_1₁(X), 0_0, []_0) -> int_3_out_gga₃(s_1₁(X), 0_0, []_0)
int_3_in_gga₃(s_1₁(X), s_1₁(Y), XS) -> if_int_3_in_2_gga₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
if_int_3_in_2_gga₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
intlist_2_in_ga₂([]_0, []_0) -> intlist_2_out_ga₂([]_0, []_0)
intlist_2_in_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_out_ga₂(XS, YS)) -> intlist_2_out_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS))
if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_out_ga₂(ZS, XS)) -> int_3_out_gga₃(s_1₁(X), s_1₁(Y), XS)
if_int_3_in_1_gga₃(Y, XS, int_3_out_gga₃(s_1₁(0_0), s_1₁(Y), XS)) -> int_3_out_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

INTLIST_2_IN_GA₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> INTLIST_2_IN_GA₂(XS, YS)

The TRS R consists of the following rules:

int_3_in_gga₃(0_0, 0_0, ._2₂(0_0, []_0)) -> int_3_out_gga₃(0_0, 0_0, ._2₂(0_0, []_0))
int_3_in_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> if_int_3_in_1_gga₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
int_3_in_gga₃(s_1₁(X), 0_0, []_0) -> int_3_out_gga₃(s_1₁(X), 0_0, []_0)
int_3_in_gga₃(s_1₁(X), s_1₁(Y), XS) -> if_int_3_in_2_gga₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
if_int_3_in_2_gga₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
intlist_2_in_ga₂([]_0, []_0) -> intlist_2_out_ga₂([]_0, []_0)
intlist_2_in_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_out_ga₂(XS, YS)) -> intlist_2_out_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS))
if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_out_ga₂(ZS, XS)) -> int_3_out_gga₃(s_1₁(X), s_1₁(Y), XS)
if_int_3_in_1_gga₃(Y, XS, int_3_out_gga₃(s_1₁(0_0), s_1₁(Y), XS)) -> int_3_out_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS))

The argument filtering Pi contains the following mapping:
int_3_in_gga₃(x1, x2, x3) = int_3_in_gga₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
int_3_out_gga₃(x1, x2, x3) = int_3_out_gga₁(x3)
if_int_3_in_1_gga₃(x1, x2, x3) = if_int_3_in_1_gga₁(x3)
if_int_3_in_2_gga₄(x1, x2, x3, x4) = if_int_3_in_2_gga₁(x4)
if_int_3_in_3_gga₅(x1, x2, x3, x4, x5) = if_int_3_in_3_gga₁(x5)
intlist_2_in_ga₂(x1, x2) = intlist_2_in_ga₁(x1)
intlist_2_out_ga₂(x1, x2) = intlist_2_out_ga₁(x2)
if_intlist_2_in_1_ga₄(x1, x2, x3, x4) = if_intlist_2_in_1_ga₂(x1, x4)
INTLIST_2_IN_GA₂(x1, x2) = INTLIST_2_IN_GA₁(x1)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

INTLIST_2_IN_GA₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> INTLIST_2_IN_GA₂(XS, YS)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
INTLIST_2_IN_GA₂(x1, x2) = INTLIST_2_IN_GA₁(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

INTLIST_2_IN_GA₁(._2₂(X, XS)) -> INTLIST_2_IN_GA₁(XS)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {INTLIST_2_IN_GA₁}.
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:

INTLIST_2_IN_GA₁(._2₂(X, XS)) -> INTLIST_2_IN_GA₁(XS)
The graph contains the following edges 1 > 1

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

INT_3_IN_GGA₃(s_1₁(X), s_1₁(Y), XS) -> INT_3_IN_GGA₃(X, Y, ZS)
INT_3_IN_GGA₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> INT_3_IN_GGA₃(s_1₁(0_0), s_1₁(Y), XS)

The TRS R consists of the following rules:

int_3_in_gga₃(0_0, 0_0, ._2₂(0_0, []_0)) -> int_3_out_gga₃(0_0, 0_0, ._2₂(0_0, []_0))
int_3_in_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> if_int_3_in_1_gga₃(Y, XS, int_3_in_gga₃(s_1₁(0_0), s_1₁(Y), XS))
int_3_in_gga₃(s_1₁(X), 0_0, []_0) -> int_3_out_gga₃(s_1₁(X), 0_0, []_0)
int_3_in_gga₃(s_1₁(X), s_1₁(Y), XS) -> if_int_3_in_2_gga₄(X, Y, XS, int_3_in_gga₃(X, Y, ZS))
if_int_3_in_2_gga₄(X, Y, XS, int_3_out_gga₃(X, Y, ZS)) -> if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_in_ga₂(ZS, XS))
intlist_2_in_ga₂([]_0, []_0) -> intlist_2_out_ga₂([]_0, []_0)
intlist_2_in_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS)) -> if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_in_ga₂(XS, YS))
if_intlist_2_in_1_ga₄(X, XS, YS, intlist_2_out_ga₂(XS, YS)) -> intlist_2_out_ga₂(._2₂(X, XS), ._2₂(s_1₁(X), YS))
if_int_3_in_3_gga₅(X, Y, XS, ZS, intlist_2_out_ga₂(ZS, XS)) -> int_3_out_gga₃(s_1₁(X), s_1₁(Y), XS)
if_int_3_in_1_gga₃(Y, XS, int_3_out_gga₃(s_1₁(0_0), s_1₁(Y), XS)) -> int_3_out_gga₃(0_0, s_1₁(Y), ._2₂(0_0, XS))

The argument filtering Pi contains the following mapping:
int_3_in_gga₃(x1, x2, x3) = int_3_in_gga₂(x1, x2)
[]_0 = []_0
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
int_3_out_gga₃(x1, x2, x3) = int_3_out_gga₁(x3)
if_int_3_in_1_gga₃(x1, x2, x3) = if_int_3_in_1_gga₁(x3)
if_int_3_in_2_gga₄(x1, x2, x3, x4) = if_int_3_in_2_gga₁(x4)
if_int_3_in_3_gga₅(x1, x2, x3, x4, x5) = if_int_3_in_3_gga₁(x5)
intlist_2_in_ga₂(x1, x2) = intlist_2_in_ga₁(x1)
intlist_2_out_ga₂(x1, x2) = intlist_2_out_ga₁(x2)
if_intlist_2_in_1_ga₄(x1, x2, x3, x4) = if_intlist_2_in_1_ga₂(x1, x4)
INT_3_IN_GGA₃(x1, x2, x3) = INT_3_IN_GGA₂(x1, x2)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

INT_3_IN_GGA₃(s_1₁(X), s_1₁(Y), XS) -> INT_3_IN_GGA₃(X, Y, ZS)
INT_3_IN_GGA₃(0_0, s_1₁(Y), ._2₂(0_0, XS)) -> INT_3_IN_GGA₃(s_1₁(0_0), s_1₁(Y), XS)

R is empty.
The argument filtering Pi contains the following mapping:
._2₂(x1, x2) = ._2₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
INT_3_IN_GGA₃(x1, x2, x3) = INT_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

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

INT_3_IN_GGA₂(s_1₁(X), s_1₁(Y)) -> INT_3_IN_GGA₂(X, Y)
INT_3_IN_GGA₂(0_0, s_1₁(Y)) -> INT_3_IN_GGA₂(s_1₁(0_0), s_1₁(Y))

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

INT_3_IN_GGA₂(s_1₁(X), s_1₁(Y)) -> INT_3_IN_GGA₂(X, Y)
The graph contains the following edges 1 > 1, 2 > 2
INT_3_IN_GGA₂(0_0, s_1₁(Y)) -> INT_3_IN_GGA₂(s_1₁(0_0), s_1₁(Y))
The graph contains the following edges 2 >= 2