Left Termination proof of /tmp/tmpAwCTzi/pplus2.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

p₂(s₁(0₀), 0₀).
p₂(s₁²(X), s₁²(Y)) :- p₂(s₁(X), s₁(Y)).
plus₃(0₀, Y, Y).
plus₃(s₁(X), Y, s₁(Z)) :- p₂(s₁(X), U), plus₃(U, Y, Z).

With regard to the inferred argument filtering the predicates were used in the following modes:
plus₃: (b,f,f)
p₂: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_3_in_gaa₃(0_0, Y, Y) -> plus_3_out_gaa₃(0_0, Y, Y)
plus_3_in_gaa₃(s_1₁(X), Y, s_1₁(Z)) -> if_plus_3_in_1_gaa₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))
if_plus_3_in_1_gaa₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_out_gaa₃(U, Y, Z)) -> plus_3_out_gaa₃(s_1₁(X), Y, s_1₁(Z))

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:

plus_3_in_gaa₃(0_0, Y, Y) -> plus_3_out_gaa₃(0_0, Y, Y)
plus_3_in_gaa₃(s_1₁(X), Y, s_1₁(Z)) -> if_plus_3_in_1_gaa₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))
if_plus_3_in_1_gaa₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_out_gaa₃(U, Y, Z)) -> plus_3_out_gaa₃(s_1₁(X), Y, s_1₁(Z))

PLUS_3_IN_GAA₃(s_1₁(X), Y, s_1₁(Z)) -> IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
PLUS_3_IN_GAA₃(s_1₁(X), Y, s_1₁(Z)) -> P_2_IN_GA₂(s_1₁(X), U)
P_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> IF_P_2_IN_1_GA₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
P_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> P_2_IN_GA₂(s_1₁(X), s_1₁(Y))
IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> IF_PLUS_3_IN_2_GAA₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> PLUS_3_IN_GAA₃(U, Y, Z)

The TRS R consists of the following rules:

plus_3_in_gaa₃(0_0, Y, Y) -> plus_3_out_gaa₃(0_0, Y, Y)
plus_3_in_gaa₃(s_1₁(X), Y, s_1₁(Z)) -> if_plus_3_in_1_gaa₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))
if_plus_3_in_1_gaa₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_out_gaa₃(U, Y, Z)) -> plus_3_out_gaa₃(s_1₁(X), Y, s_1₁(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa₃(x1, x2, x3) = plus_3_in_gaa₁(x1)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
plus_3_out_gaa₃(x1, x2, x3) = plus_3_out_gaa
if_plus_3_in_1_gaa₄(x1, x2, x3, x4) = if_plus_3_in_1_gaa₁(x4)
p_2_in_ga₂(x1, x2) = p_2_in_ga₁(x1)
p_2_out_ga₂(x1, x2) = p_2_out_ga₁(x2)
if_p_2_in_1_ga₃(x1, x2, x3) = if_p_2_in_1_ga₁(x3)
if_plus_3_in_2_gaa₅(x1, x2, x3, x4, x5) = if_plus_3_in_2_gaa₁(x5)
IF_PLUS_3_IN_1_GAA₄(x1, x2, x3, x4) = IF_PLUS_3_IN_1_GAA₁(x4)
IF_P_2_IN_1_GA₃(x1, x2, x3) = IF_P_2_IN_1_GA₁(x3)
P_2_IN_GA₂(x1, x2) = P_2_IN_GA₁(x1)
PLUS_3_IN_GAA₃(x1, x2, x3) = PLUS_3_IN_GAA₁(x1)
IF_PLUS_3_IN_2_GAA₅(x1, x2, x3, x4, x5) = IF_PLUS_3_IN_2_GAA₁(x5)

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:

PLUS_3_IN_GAA₃(s_1₁(X), Y, s_1₁(Z)) -> IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
PLUS_3_IN_GAA₃(s_1₁(X), Y, s_1₁(Z)) -> P_2_IN_GA₂(s_1₁(X), U)
P_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> IF_P_2_IN_1_GA₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
P_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> P_2_IN_GA₂(s_1₁(X), s_1₁(Y))
IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> IF_PLUS_3_IN_2_GAA₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> PLUS_3_IN_GAA₃(U, Y, Z)

The TRS R consists of the following rules:

plus_3_in_gaa₃(0_0, Y, Y) -> plus_3_out_gaa₃(0_0, Y, Y)
plus_3_in_gaa₃(s_1₁(X), Y, s_1₁(Z)) -> if_plus_3_in_1_gaa₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))
if_plus_3_in_1_gaa₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_out_gaa₃(U, Y, Z)) -> plus_3_out_gaa₃(s_1₁(X), Y, s_1₁(Z))

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

P_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> P_2_IN_GA₂(s_1₁(X), s_1₁(Y))

The TRS R consists of the following rules:

plus_3_in_gaa₃(0_0, Y, Y) -> plus_3_out_gaa₃(0_0, Y, Y)
plus_3_in_gaa₃(s_1₁(X), Y, s_1₁(Z)) -> if_plus_3_in_1_gaa₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))
if_plus_3_in_1_gaa₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_out_gaa₃(U, Y, Z)) -> plus_3_out_gaa₃(s_1₁(X), Y, s_1₁(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa₃(x1, x2, x3) = plus_3_in_gaa₁(x1)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
plus_3_out_gaa₃(x1, x2, x3) = plus_3_out_gaa
if_plus_3_in_1_gaa₄(x1, x2, x3, x4) = if_plus_3_in_1_gaa₁(x4)
p_2_in_ga₂(x1, x2) = p_2_in_ga₁(x1)
p_2_out_ga₂(x1, x2) = p_2_out_ga₁(x2)
if_p_2_in_1_ga₃(x1, x2, x3) = if_p_2_in_1_ga₁(x3)
if_plus_3_in_2_gaa₅(x1, x2, x3, x4, x5) = if_plus_3_in_2_gaa₁(x5)
P_2_IN_GA₂(x1, x2) = P_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:

P_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> P_2_IN_GA₂(s_1₁(X), s_1₁(Y))

R is empty.
The argument filtering Pi contains the following mapping:
s_1₁(x1) = s_1₁(x1)
P_2_IN_GA₂(x1, x2) = P_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:

P_2_IN_GA₁(s_1₁(s_1₁(X))) -> P_2_IN_GA₁(s_1₁(X))

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

P_2_IN_GA₁(s_1₁(s_1₁(X))) -> P_2_IN_GA₁(s_1₁(X))
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:

IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> PLUS_3_IN_GAA₃(U, Y, Z)
PLUS_3_IN_GAA₃(s_1₁(X), Y, s_1₁(Z)) -> IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))

The TRS R consists of the following rules:

plus_3_in_gaa₃(0_0, Y, Y) -> plus_3_out_gaa₃(0_0, Y, Y)
plus_3_in_gaa₃(s_1₁(X), Y, s_1₁(Z)) -> if_plus_3_in_1_gaa₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))
p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))
if_plus_3_in_1_gaa₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_in_gaa₃(U, Y, Z))
if_plus_3_in_2_gaa₅(X, Y, Z, U, plus_3_out_gaa₃(U, Y, Z)) -> plus_3_out_gaa₃(s_1₁(X), Y, s_1₁(Z))

The argument filtering Pi contains the following mapping:
plus_3_in_gaa₃(x1, x2, x3) = plus_3_in_gaa₁(x1)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
plus_3_out_gaa₃(x1, x2, x3) = plus_3_out_gaa
if_plus_3_in_1_gaa₄(x1, x2, x3, x4) = if_plus_3_in_1_gaa₁(x4)
p_2_in_ga₂(x1, x2) = p_2_in_ga₁(x1)
p_2_out_ga₂(x1, x2) = p_2_out_ga₁(x2)
if_p_2_in_1_ga₃(x1, x2, x3) = if_p_2_in_1_ga₁(x3)
if_plus_3_in_2_gaa₅(x1, x2, x3, x4, x5) = if_plus_3_in_2_gaa₁(x5)
IF_PLUS_3_IN_1_GAA₄(x1, x2, x3, x4) = IF_PLUS_3_IN_1_GAA₁(x4)
PLUS_3_IN_GAA₃(x1, x2, x3) = PLUS_3_IN_GAA₁(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_out_ga₂(s_1₁(X), U)) -> PLUS_3_IN_GAA₃(U, Y, Z)
PLUS_3_IN_GAA₃(s_1₁(X), Y, s_1₁(Z)) -> IF_PLUS_3_IN_1_GAA₄(X, Y, Z, p_2_in_ga₂(s_1₁(X), U))

The TRS R consists of the following rules:

p_2_in_ga₂(s_1₁(0_0), 0_0) -> p_2_out_ga₂(s_1₁(0_0), 0_0)
p_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y))) -> if_p_2_in_1_ga₃(X, Y, p_2_in_ga₂(s_1₁(X), s_1₁(Y)))
if_p_2_in_1_ga₃(X, Y, p_2_out_ga₂(s_1₁(X), s_1₁(Y))) -> p_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(s_1₁(Y)))

The argument filtering Pi contains the following mapping:
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
p_2_in_ga₂(x1, x2) = p_2_in_ga₁(x1)
p_2_out_ga₂(x1, x2) = p_2_out_ga₁(x2)
if_p_2_in_1_ga₃(x1, x2, x3) = if_p_2_in_1_ga₁(x3)
IF_PLUS_3_IN_1_GAA₄(x1, x2, x3, x4) = IF_PLUS_3_IN_1_GAA₁(x4)
PLUS_3_IN_GAA₃(x1, x2, x3) = PLUS_3_IN_GAA₁(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

IF_PLUS_3_IN_1_GAA₁(p_2_out_ga₁(U)) -> PLUS_3_IN_GAA₁(U)
PLUS_3_IN_GAA₁(s_1₁(X)) -> IF_PLUS_3_IN_1_GAA₁(p_2_in_ga₁(s_1₁(X)))

The TRS R consists of the following rules:

p_2_in_ga₁(s_1₁(0_0)) -> p_2_out_ga₁(0_0)
p_2_in_ga₁(s_1₁(s_1₁(X))) -> if_p_2_in_1_ga₁(p_2_in_ga₁(s_1₁(X)))
if_p_2_in_1_ga₁(p_2_out_ga₁(s_1₁(Y))) -> p_2_out_ga₁(s_1₁(s_1₁(Y)))

The set Q consists of the following terms:

p_2_in_ga₁(x0)
if_p_2_in_1_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {PLUS_3_IN_GAA₁, IF_PLUS_3_IN_1_GAA₁}.
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 dependency pairs:

IF_PLUS_3_IN_1_GAA₁(p_2_out_ga₁(U)) -> PLUS_3_IN_GAA₁(U)

Strictly oriented rules of the TRS R:

p_2_in_ga₁(s_1₁(0_0)) -> p_2_out_ga₁(0_0)
if_p_2_in_1_ga₁(p_2_out_ga₁(s_1₁(Y))) -> p_2_out_ga₁(s_1₁(s_1₁(Y)))

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 2
POL(p_2_out_ga₁(x₁)) = 1 + x₁
POL(if_p_2_in_1_ga₁(x₁)) = 2·x₁
POL(p_2_in_ga₁(x₁)) = x₁
POL(s_1₁(x₁)) = 2·x₁
POL(IF_PLUS_3_IN_1_GAA₁(x₁)) = x₁
POL(PLUS_3_IN_GAA₁(x₁)) = x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

PLUS_3_IN_GAA₁(s_1₁(X)) -> IF_PLUS_3_IN_1_GAA₁(p_2_in_ga₁(s_1₁(X)))

The TRS R consists of the following rules:

p_2_in_ga₁(s_1₁(s_1₁(X))) -> if_p_2_in_1_ga₁(p_2_in_ga₁(s_1₁(X)))

The set Q consists of the following terms:

p_2_in_ga₁(x0)
if_p_2_in_1_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PLUS_3_IN_1_GAA₁, PLUS_3_IN_GAA₁}.
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:

p_2_in_ga₁(s_1₁(s_1₁(X))) -> if_p_2_in_1_ga₁(p_2_in_ga₁(s_1₁(X)))

Used ordering: POLO with Polynomial interpretation:

POL(if_p_2_in_1_ga₁(x₁)) = 1 + x₁
POL(p_2_in_ga₁(x₁)) = 1 + x₁
POL(s_1₁(x₁)) = 1 + 2·x₁
POL(IF_PLUS_3_IN_1_GAA₁(x₁)) = x₁
POL(PLUS_3_IN_GAA₁(x₁)) = 2·x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

PLUS_3_IN_GAA₁(s_1₁(X)) -> IF_PLUS_3_IN_1_GAA₁(p_2_in_ga₁(s_1₁(X)))

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

p_2_in_ga₁(x0)
if_p_2_in_1_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_PLUS_3_IN_1_GAA₁, PLUS_3_IN_GAA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.