Left Termination proof of /tmp/tmpdyJQO-/log2a.pl

Left Termination of the query pattern log2(b,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

log2₂(X, Y) :- log2₃(X, 0₀, Y).
log2₃(0₀, I, I).
log2₃(s₁(0₀), I, I).
log2₃(s₁²(X), I, Y) :- half₂(s₁²(X), X1), log2₃(X1, s₁(I), Y).
half₂(0₀, 0₀).
half₂(s₁(0₀), 0₀).
half₂(s₁²(X), s₁(Y)) :- half₂(X, Y).

With regard to the inferred argument filtering the predicates were used in the following modes:
log2₂: (b,f)
log2₃: (b,b,f)
half₂: (b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

log2_2_in_ga₂(X, Y) -> if_log2_2_in_1_ga₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
log2_3_in_gga₃(0_0, I, I) -> log2_3_out_gga₃(0_0, I, I)
log2_3_in_gga₃(s_1₁(0_0), I, I) -> log2_3_out_gga₃(s_1₁(0_0), I, I)
log2_3_in_gga₃(s_1₁(s_1₁(X)), I, Y) -> if_log2_3_in_1_gga₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)
half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
if_log2_3_in_1_gga₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_out_gga₃(X1, s_1₁(I), Y)) -> log2_3_out_gga₃(s_1₁(s_1₁(X)), I, Y)
if_log2_2_in_1_ga₃(X, Y, log2_3_out_gga₃(X, 0_0, Y)) -> log2_2_out_ga₂(X, Y)

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:

log2_2_in_ga₂(X, Y) -> if_log2_2_in_1_ga₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
log2_3_in_gga₃(0_0, I, I) -> log2_3_out_gga₃(0_0, I, I)
log2_3_in_gga₃(s_1₁(0_0), I, I) -> log2_3_out_gga₃(s_1₁(0_0), I, I)
log2_3_in_gga₃(s_1₁(s_1₁(X)), I, Y) -> if_log2_3_in_1_gga₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)
half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
if_log2_3_in_1_gga₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_out_gga₃(X1, s_1₁(I), Y)) -> log2_3_out_gga₃(s_1₁(s_1₁(X)), I, Y)
if_log2_2_in_1_ga₃(X, Y, log2_3_out_gga₃(X, 0_0, Y)) -> log2_2_out_ga₂(X, Y)

LOG2_2_IN_GA₂(X, Y) -> IF_LOG2_2_IN_1_GA₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
LOG2_2_IN_GA₂(X, Y) -> LOG2_3_IN_GGA₃(X, 0_0, Y)
LOG2_3_IN_GGA₃(s_1₁(s_1₁(X)), I, Y) -> IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
LOG2_3_IN_GGA₃(s_1₁(s_1₁(X)), I, Y) -> HALF_2_IN_GA₂(s_1₁(s_1₁(X)), X1)
HALF_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> IF_HALF_2_IN_1_GA₃(X, Y, half_2_in_ga₂(X, Y))
HALF_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> HALF_2_IN_GA₂(X, Y)
IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> IF_LOG2_3_IN_2_GGA₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> LOG2_3_IN_GGA₃(X1, s_1₁(I), Y)

The TRS R consists of the following rules:

log2_2_in_ga₂(X, Y) -> if_log2_2_in_1_ga₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
log2_3_in_gga₃(0_0, I, I) -> log2_3_out_gga₃(0_0, I, I)
log2_3_in_gga₃(s_1₁(0_0), I, I) -> log2_3_out_gga₃(s_1₁(0_0), I, I)
log2_3_in_gga₃(s_1₁(s_1₁(X)), I, Y) -> if_log2_3_in_1_gga₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)
half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
if_log2_3_in_1_gga₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_out_gga₃(X1, s_1₁(I), Y)) -> log2_3_out_gga₃(s_1₁(s_1₁(X)), I, Y)
if_log2_2_in_1_ga₃(X, Y, log2_3_out_gga₃(X, 0_0, Y)) -> log2_2_out_ga₂(X, Y)

The argument filtering Pi contains the following mapping:
log2_2_in_ga₂(x1, x2) = log2_2_in_ga₁(x1)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
if_log2_2_in_1_ga₃(x1, x2, x3) = if_log2_2_in_1_ga₁(x3)
log2_3_in_gga₃(x1, x2, x3) = log2_3_in_gga₂(x1, x2)
log2_3_out_gga₃(x1, x2, x3) = log2_3_out_gga₁(x3)
if_log2_3_in_1_gga₄(x1, x2, x3, x4) = if_log2_3_in_1_gga₂(x2, x4)
half_2_in_ga₂(x1, x2) = half_2_in_ga₁(x1)
half_2_out_ga₂(x1, x2) = half_2_out_ga₁(x2)
if_half_2_in_1_ga₃(x1, x2, x3) = if_half_2_in_1_ga₁(x3)
if_log2_3_in_2_gga₅(x1, x2, x3, x4, x5) = if_log2_3_in_2_gga₁(x5)
log2_2_out_ga₂(x1, x2) = log2_2_out_ga₁(x2)
IF_LOG2_2_IN_1_GA₃(x1, x2, x3) = IF_LOG2_2_IN_1_GA₁(x3)
HALF_2_IN_GA₂(x1, x2) = HALF_2_IN_GA₁(x1)
LOG2_3_IN_GGA₃(x1, x2, x3) = LOG2_3_IN_GGA₂(x1, x2)
IF_LOG2_3_IN_1_GGA₄(x1, x2, x3, x4) = IF_LOG2_3_IN_1_GGA₂(x2, x4)
IF_HALF_2_IN_1_GA₃(x1, x2, x3) = IF_HALF_2_IN_1_GA₁(x3)
LOG2_2_IN_GA₂(x1, x2) = LOG2_2_IN_GA₁(x1)
IF_LOG2_3_IN_2_GGA₅(x1, x2, x3, x4, x5) = IF_LOG2_3_IN_2_GGA₁(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:

LOG2_2_IN_GA₂(X, Y) -> IF_LOG2_2_IN_1_GA₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
LOG2_2_IN_GA₂(X, Y) -> LOG2_3_IN_GGA₃(X, 0_0, Y)
LOG2_3_IN_GGA₃(s_1₁(s_1₁(X)), I, Y) -> IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
LOG2_3_IN_GGA₃(s_1₁(s_1₁(X)), I, Y) -> HALF_2_IN_GA₂(s_1₁(s_1₁(X)), X1)
HALF_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> IF_HALF_2_IN_1_GA₃(X, Y, half_2_in_ga₂(X, Y))
HALF_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> HALF_2_IN_GA₂(X, Y)
IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> IF_LOG2_3_IN_2_GGA₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> LOG2_3_IN_GGA₃(X1, s_1₁(I), Y)

The TRS R consists of the following rules:

log2_2_in_ga₂(X, Y) -> if_log2_2_in_1_ga₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
log2_3_in_gga₃(0_0, I, I) -> log2_3_out_gga₃(0_0, I, I)
log2_3_in_gga₃(s_1₁(0_0), I, I) -> log2_3_out_gga₃(s_1₁(0_0), I, I)
log2_3_in_gga₃(s_1₁(s_1₁(X)), I, Y) -> if_log2_3_in_1_gga₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)
half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
if_log2_3_in_1_gga₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_out_gga₃(X1, s_1₁(I), Y)) -> log2_3_out_gga₃(s_1₁(s_1₁(X)), I, Y)
if_log2_2_in_1_ga₃(X, Y, log2_3_out_gga₃(X, 0_0, Y)) -> log2_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

HALF_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> HALF_2_IN_GA₂(X, Y)

The TRS R consists of the following rules:

log2_2_in_ga₂(X, Y) -> if_log2_2_in_1_ga₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
log2_3_in_gga₃(0_0, I, I) -> log2_3_out_gga₃(0_0, I, I)
log2_3_in_gga₃(s_1₁(0_0), I, I) -> log2_3_out_gga₃(s_1₁(0_0), I, I)
log2_3_in_gga₃(s_1₁(s_1₁(X)), I, Y) -> if_log2_3_in_1_gga₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)
half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
if_log2_3_in_1_gga₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_out_gga₃(X1, s_1₁(I), Y)) -> log2_3_out_gga₃(s_1₁(s_1₁(X)), I, Y)
if_log2_2_in_1_ga₃(X, Y, log2_3_out_gga₃(X, 0_0, Y)) -> log2_2_out_ga₂(X, Y)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

HALF_2_IN_GA₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> HALF_2_IN_GA₂(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s_1₁(x1) = s_1₁(x1)
HALF_2_IN_GA₂(x1, x2) = HALF_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:

HALF_2_IN_GA₁(s_1₁(s_1₁(X))) -> HALF_2_IN_GA₁(X)

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

HALF_2_IN_GA₁(s_1₁(s_1₁(X))) -> HALF_2_IN_GA₁(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:

LOG2_3_IN_GGA₃(s_1₁(s_1₁(X)), I, Y) -> IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> LOG2_3_IN_GGA₃(X1, s_1₁(I), Y)

The TRS R consists of the following rules:

log2_2_in_ga₂(X, Y) -> if_log2_2_in_1_ga₃(X, Y, log2_3_in_gga₃(X, 0_0, Y))
log2_3_in_gga₃(0_0, I, I) -> log2_3_out_gga₃(0_0, I, I)
log2_3_in_gga₃(s_1₁(0_0), I, I) -> log2_3_out_gga₃(s_1₁(0_0), I, I)
log2_3_in_gga₃(s_1₁(s_1₁(X)), I, Y) -> if_log2_3_in_1_gga₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)
half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
if_log2_3_in_1_gga₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_in_gga₃(X1, s_1₁(I), Y))
if_log2_3_in_2_gga₅(X, I, Y, X1, log2_3_out_gga₃(X1, s_1₁(I), Y)) -> log2_3_out_gga₃(s_1₁(s_1₁(X)), I, Y)
if_log2_2_in_1_ga₃(X, Y, log2_3_out_gga₃(X, 0_0, Y)) -> log2_2_out_ga₂(X, Y)

The argument filtering Pi contains the following mapping:
log2_2_in_ga₂(x1, x2) = log2_2_in_ga₁(x1)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
if_log2_2_in_1_ga₃(x1, x2, x3) = if_log2_2_in_1_ga₁(x3)
log2_3_in_gga₃(x1, x2, x3) = log2_3_in_gga₂(x1, x2)
log2_3_out_gga₃(x1, x2, x3) = log2_3_out_gga₁(x3)
if_log2_3_in_1_gga₄(x1, x2, x3, x4) = if_log2_3_in_1_gga₂(x2, x4)
half_2_in_ga₂(x1, x2) = half_2_in_ga₁(x1)
half_2_out_ga₂(x1, x2) = half_2_out_ga₁(x2)
if_half_2_in_1_ga₃(x1, x2, x3) = if_half_2_in_1_ga₁(x3)
if_log2_3_in_2_gga₅(x1, x2, x3, x4, x5) = if_log2_3_in_2_gga₁(x5)
log2_2_out_ga₂(x1, x2) = log2_2_out_ga₁(x2)
LOG2_3_IN_GGA₃(x1, x2, x3) = LOG2_3_IN_GGA₂(x1, x2)
IF_LOG2_3_IN_1_GGA₄(x1, x2, x3, x4) = IF_LOG2_3_IN_1_GGA₂(x2, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

LOG2_3_IN_GGA₃(s_1₁(s_1₁(X)), I, Y) -> IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_in_ga₂(s_1₁(s_1₁(X)), X1))
IF_LOG2_3_IN_1_GGA₄(X, I, Y, half_2_out_ga₂(s_1₁(s_1₁(X)), X1)) -> LOG2_3_IN_GGA₃(X1, s_1₁(I), Y)

The TRS R consists of the following rules:

half_2_in_ga₂(s_1₁(s_1₁(X)), s_1₁(Y)) -> if_half_2_in_1_ga₃(X, Y, half_2_in_ga₂(X, Y))
if_half_2_in_1_ga₃(X, Y, half_2_out_ga₂(X, Y)) -> half_2_out_ga₂(s_1₁(s_1₁(X)), s_1₁(Y))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(0_0, 0_0)
half_2_in_ga₂(s_1₁(0_0), 0_0) -> half_2_out_ga₂(s_1₁(0_0), 0_0)

The argument filtering Pi contains the following mapping:
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
half_2_in_ga₂(x1, x2) = half_2_in_ga₁(x1)
half_2_out_ga₂(x1, x2) = half_2_out_ga₁(x2)
if_half_2_in_1_ga₃(x1, x2, x3) = if_half_2_in_1_ga₁(x3)
LOG2_3_IN_GGA₃(x1, x2, x3) = LOG2_3_IN_GGA₂(x1, x2)
IF_LOG2_3_IN_1_GGA₄(x1, x2, x3, x4) = IF_LOG2_3_IN_1_GGA₂(x2, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

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

LOG2_3_IN_GGA₂(s_1₁(s_1₁(X)), I) -> IF_LOG2_3_IN_1_GGA₂(I, half_2_in_ga₁(s_1₁(s_1₁(X))))
IF_LOG2_3_IN_1_GGA₂(I, half_2_out_ga₁(X1)) -> LOG2_3_IN_GGA₂(X1, s_1₁(I))

The TRS R consists of the following rules:

half_2_in_ga₁(s_1₁(s_1₁(X))) -> if_half_2_in_1_ga₁(half_2_in_ga₁(X))
if_half_2_in_1_ga₁(half_2_out_ga₁(Y)) -> half_2_out_ga₁(s_1₁(Y))
half_2_in_ga₁(0_0) -> half_2_out_ga₁(0_0)
half_2_in_ga₁(s_1₁(0_0)) -> half_2_out_ga₁(0_0)

The set Q consists of the following terms:

half_2_in_ga₁(x0)
if_half_2_in_1_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {IF_LOG2_3_IN_1_GGA₂, LOG2_3_IN_GGA₂}.
By using a polynomial ordering, the following set of Dependency Pairs of this DP problem can be strictly oriented.

LOG2_3_IN_GGA₂(s_1₁(s_1₁(X)), I) -> IF_LOG2_3_IN_1_GGA₂(I, half_2_in_ga₁(s_1₁(s_1₁(X))))

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

IF_LOG2_3_IN_1_GGA₂(I, half_2_out_ga₁(X1)) -> LOG2_3_IN_GGA₂(X1, s_1₁(I))

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

half_2_in_ga₁(s_1₁(s_1₁(X))) -> if_half_2_in_1_ga₁(half_2_in_ga₁(X))
half_2_in_ga₁(0_0) -> half_2_out_ga₁(0_0)
if_half_2_in_1_ga₁(half_2_out_ga₁(Y)) -> half_2_out_ga₁(s_1₁(Y))
half_2_in_ga₁(s_1₁(0_0)) -> half_2_out_ga₁(0_0)

Used ordering: POLO with Polynomial interpretation:

POL(if_half_2_in_1_ga₁(x₁)) = 2 + x₁
POL(0_0) = 0
POL(half_2_in_ga₁(x₁)) = x₁
POL(LOG2_3_IN_GGA₂(x₁, x₂)) = 2·x₁
POL(half_2_out_ga₁(x₁)) = 2·x₁
POL(IF_LOG2_3_IN_1_GGA₂(x₁, x₂)) = x₂
POL(s_1₁(x₁)) = 1 + x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPPoloProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF_LOG2_3_IN_1_GGA₂(I, half_2_out_ga₁(X1)) -> LOG2_3_IN_GGA₂(X1, s_1₁(I))

The TRS R consists of the following rules:

half_2_in_ga₁(s_1₁(s_1₁(X))) -> if_half_2_in_1_ga₁(half_2_in_ga₁(X))
if_half_2_in_1_ga₁(half_2_out_ga₁(Y)) -> half_2_out_ga₁(s_1₁(Y))
half_2_in_ga₁(0_0) -> half_2_out_ga₁(0_0)
half_2_in_ga₁(s_1₁(0_0)) -> half_2_out_ga₁(0_0)

The set Q consists of the following terms:

half_2_in_ga₁(x0)
if_half_2_in_1_ga₁(x0)

We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {LOG2_3_IN_GGA₂, IF_LOG2_3_IN_1_GGA₂}.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.