Left Termination proof of /tmp/tmppObCZc/log.pl

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

↳ PROLOG

  ↳ PrologToPiTRSProof

half₂(0₀, 0₀).
half₂(s₁(0₀), 0₀).
half₂(s₁²(X), s₁(Y)) :- half₂(X, Y).
log₂(0₀, s₁(0₀)).
log₂(s₁(X), s₁(Y)) :- half₂(s₁(X), Z), log₂(Z, Y).

With regard to the inferred argument filtering the predicates were used in the following modes:
log₂: (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:

log_2_in_ga₂(0_0, s_1₁(0_0)) -> log_2_out_ga₂(0_0, s_1₁(0_0))
log_2_in_ga₂(s_1₁(X), s_1₁(Y)) -> if_log_2_in_1_ga₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
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_log_2_in_1_ga₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> if_log_2_in_2_ga₄(X, Y, Z, log_2_in_ga₂(Z, Y))
if_log_2_in_2_ga₄(X, Y, Z, log_2_out_ga₂(Z, Y)) -> log_2_out_ga₂(s_1₁(X), s_1₁(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:

log_2_in_ga₂(0_0, s_1₁(0_0)) -> log_2_out_ga₂(0_0, s_1₁(0_0))
log_2_in_ga₂(s_1₁(X), s_1₁(Y)) -> if_log_2_in_1_ga₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
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_log_2_in_1_ga₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> if_log_2_in_2_ga₄(X, Y, Z, log_2_in_ga₂(Z, Y))
if_log_2_in_2_ga₄(X, Y, Z, log_2_out_ga₂(Z, Y)) -> log_2_out_ga₂(s_1₁(X), s_1₁(Y))

LOG_2_IN_GA₂(s_1₁(X), s_1₁(Y)) -> IF_LOG_2_IN_1_GA₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
LOG_2_IN_GA₂(s_1₁(X), s_1₁(Y)) -> HALF_2_IN_GA₂(s_1₁(X), Z)
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_LOG_2_IN_1_GA₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> IF_LOG_2_IN_2_GA₄(X, Y, Z, log_2_in_ga₂(Z, Y))
IF_LOG_2_IN_1_GA₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> LOG_2_IN_GA₂(Z, Y)

The TRS R consists of the following rules:

log_2_in_ga₂(0_0, s_1₁(0_0)) -> log_2_out_ga₂(0_0, s_1₁(0_0))
log_2_in_ga₂(s_1₁(X), s_1₁(Y)) -> if_log_2_in_1_ga₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
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_log_2_in_1_ga₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> if_log_2_in_2_ga₄(X, Y, Z, log_2_in_ga₂(Z, Y))
if_log_2_in_2_ga₄(X, Y, Z, log_2_out_ga₂(Z, Y)) -> log_2_out_ga₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
log_2_in_ga₂(x1, x2) = log_2_in_ga₁(x1)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
log_2_out_ga₂(x1, x2) = log_2_out_ga₁(x2)
if_log_2_in_1_ga₃(x1, x2, x3) = if_log_2_in_1_ga₁(x3)
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_log_2_in_2_ga₄(x1, x2, x3, x4) = if_log_2_in_2_ga₁(x4)
HALF_2_IN_GA₂(x1, x2) = HALF_2_IN_GA₁(x1)
IF_LOG_2_IN_1_GA₃(x1, x2, x3) = IF_LOG_2_IN_1_GA₁(x3)
LOG_2_IN_GA₂(x1, x2) = LOG_2_IN_GA₁(x1)
IF_LOG_2_IN_2_GA₄(x1, x2, x3, x4) = IF_LOG_2_IN_2_GA₁(x4)
IF_HALF_2_IN_1_GA₃(x1, x2, x3) = IF_HALF_2_IN_1_GA₁(x3)

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:

LOG_2_IN_GA₂(s_1₁(X), s_1₁(Y)) -> IF_LOG_2_IN_1_GA₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
LOG_2_IN_GA₂(s_1₁(X), s_1₁(Y)) -> HALF_2_IN_GA₂(s_1₁(X), Z)
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_LOG_2_IN_1_GA₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> IF_LOG_2_IN_2_GA₄(X, Y, Z, log_2_in_ga₂(Z, Y))
IF_LOG_2_IN_1_GA₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> LOG_2_IN_GA₂(Z, Y)

The TRS R consists of the following rules:

log_2_in_ga₂(0_0, s_1₁(0_0)) -> log_2_out_ga₂(0_0, s_1₁(0_0))
log_2_in_ga₂(s_1₁(X), s_1₁(Y)) -> if_log_2_in_1_ga₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
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_log_2_in_1_ga₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> if_log_2_in_2_ga₄(X, Y, Z, log_2_in_ga₂(Z, Y))
if_log_2_in_2_ga₄(X, Y, Z, log_2_out_ga₂(Z, Y)) -> log_2_out_ga₂(s_1₁(X), s_1₁(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:

log_2_in_ga₂(0_0, s_1₁(0_0)) -> log_2_out_ga₂(0_0, s_1₁(0_0))
log_2_in_ga₂(s_1₁(X), s_1₁(Y)) -> if_log_2_in_1_ga₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
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_log_2_in_1_ga₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> if_log_2_in_2_ga₄(X, Y, Z, log_2_in_ga₂(Z, Y))
if_log_2_in_2_ga₄(X, Y, Z, log_2_out_ga₂(Z, Y)) -> log_2_out_ga₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
log_2_in_ga₂(x1, x2) = log_2_in_ga₁(x1)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
log_2_out_ga₂(x1, x2) = log_2_out_ga₁(x2)
if_log_2_in_1_ga₃(x1, x2, x3) = if_log_2_in_1_ga₁(x3)
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_log_2_in_2_ga₄(x1, x2, x3, x4) = if_log_2_in_2_ga₁(x4)
HALF_2_IN_GA₂(x1, x2) = HALF_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:

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:

IF_LOG_2_IN_1_GA₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> LOG_2_IN_GA₂(Z, Y)
LOG_2_IN_GA₂(s_1₁(X), s_1₁(Y)) -> IF_LOG_2_IN_1_GA₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))

The TRS R consists of the following rules:

log_2_in_ga₂(0_0, s_1₁(0_0)) -> log_2_out_ga₂(0_0, s_1₁(0_0))
log_2_in_ga₂(s_1₁(X), s_1₁(Y)) -> if_log_2_in_1_ga₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))
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_log_2_in_1_ga₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> if_log_2_in_2_ga₄(X, Y, Z, log_2_in_ga₂(Z, Y))
if_log_2_in_2_ga₄(X, Y, Z, log_2_out_ga₂(Z, Y)) -> log_2_out_ga₂(s_1₁(X), s_1₁(Y))

The argument filtering Pi contains the following mapping:
log_2_in_ga₂(x1, x2) = log_2_in_ga₁(x1)
0_0 = 0_0
s_1₁(x1) = s_1₁(x1)
log_2_out_ga₂(x1, x2) = log_2_out_ga₁(x2)
if_log_2_in_1_ga₃(x1, x2, x3) = if_log_2_in_1_ga₁(x3)
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_log_2_in_2_ga₄(x1, x2, x3, x4) = if_log_2_in_2_ga₁(x4)
IF_LOG_2_IN_1_GA₃(x1, x2, x3) = IF_LOG_2_IN_1_GA₁(x3)
LOG_2_IN_GA₂(x1, x2) = LOG_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

IF_LOG_2_IN_1_GA₃(X, Y, half_2_out_ga₂(s_1₁(X), Z)) -> LOG_2_IN_GA₂(Z, Y)
LOG_2_IN_GA₂(s_1₁(X), s_1₁(Y)) -> IF_LOG_2_IN_1_GA₃(X, Y, half_2_in_ga₂(s_1₁(X), Z))

The TRS R consists of the following rules:

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))
half_2_in_ga₂(0_0, 0_0) -> half_2_out_ga₂(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)
IF_LOG_2_IN_1_GA₃(x1, x2, x3) = IF_LOG_2_IN_1_GA₁(x3)
LOG_2_IN_GA₂(x1, x2) = LOG_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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

IF_LOG_2_IN_1_GA₁(half_2_out_ga₁(Z)) -> LOG_2_IN_GA₁(Z)
LOG_2_IN_GA₁(s_1₁(X)) -> IF_LOG_2_IN_1_GA₁(half_2_in_ga₁(s_1₁(X)))

The TRS R consists of the following rules:

half_2_in_ga₁(s_1₁(0_0)) -> half_2_out_ga₁(0_0)
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)

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 {LOG_2_IN_GA₁, IF_LOG_2_IN_1_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:

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

Used ordering: POLO with Polynomial interpretation:

POL(0_0) = 1
POL(if_half_2_in_1_ga₁(x₁)) = 2·x₁
POL(LOG_2_IN_GA₁(x₁)) = 1 + x₁
POL(half_2_in_ga₁(x₁)) = 1 + x₁
POL(half_2_out_ga₁(x₁)) = 1 + x₁
POL(IF_LOG_2_IN_1_GA₁(x₁)) = x₁
POL(s_1₁(x₁)) = 1 + 2·x₁

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF_LOG_2_IN_1_GA₁(half_2_out_ga₁(Z)) -> LOG_2_IN_GA₁(Z)
LOG_2_IN_GA₁(s_1₁(X)) -> IF_LOG_2_IN_1_GA₁(half_2_in_ga₁(s_1₁(X)))

The TRS R consists of the following rules:

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)

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 {LOG_2_IN_GA₁, IF_LOG_2_IN_1_GA₁}.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.