Termination proof of /tmp/tmpLwXfHB/A12.itrs

Termination of the given ITRSProblem could successfully be proven:

↳ ITRS

  ↳ ITRStoIDPProof

ITRS problem:
The following domains are used:

z

The TRS R consists of the following rules:

f(TRUE, x, y) → f(&&(>@z(x, y), >@z(y, 2@z)), +@z(1@z, x), *@z(2@z, y))

The set Q consists of the following terms:

f(TRUE, x0, x1)

Added dependency pairs

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

f(TRUE, x, y) → f(&&(>@z(x, y), >@z(y, 2@z)), +@z(1@z, x), *@z(2@z, y))

The integer pair graph contains the following rules and edges:

(0): F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0]))

(0) -> (0), if ((+@z(1@z, x[0]) →^* x[0]a)∧(*@z(2@z, y[0]) →^* y[0]a)∧(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)) →^* TRUE))

The set Q consists of the following terms:

f(TRUE, x0, x1)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(0): F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0]))

(0) -> (0), if ((+@z(1@z, x[0]) →^* x[0]a)∧(*@z(2@z, y[0]) →^* y[0]a)∧(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)) →^* TRUE))

The set Q consists of the following terms:

f(TRUE, x0, x1)

The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.

For Pair F(TRUE, x, y) → F(&&(>@z(x, y), >@z(y, 2@z)), +@z(1@z, x), *@z(2@z, y)) the following chains were created:

We consider the chain F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0])), F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0])), F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0])) which results in the following constraint:
(1)    (+@z(1@z, x[0])=x[0]1∧&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z))=TRUE∧&&(>@z(x[0], y[0]), >@z(y[0], 2@z))=TRUE∧+@z(1@z, x[0]1)=x[0]2∧*@z(2@z, y[0])=y[0]1∧*@z(2@z, y[0]1)=y[0]2 ⇒ F(TRUE, x[0]1, y[0]1)≥NonInfC∧F(TRUE, x[0]1, y[0]1)≥F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥))

We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
(2)    (>@z(+@z(1@z, x[0]), *@z(2@z, y[0]))=TRUE∧>@z(*@z(2@z, y[0]), 2@z)=TRUE∧>@z(x[0], y[0])=TRUE∧>@z(y[0], 2@z)=TRUE ⇒ F(TRUE, +@z(1@z, x[0]), *@z(2@z, y[0]))≥NonInfC∧F(TRUE, +@z(1@z, x[0]), *@z(2@z, y[0]))≥F(&&(>@z(+@z(1@z, x[0]), *@z(2@z, y[0])), >@z(*@z(2@z, y[0]), 2@z)), +@z(1@z, +@z(1@z, x[0])), *@z(2@z, *@z(2@z, y[0])))∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3)    (x[0] + (-2)y[0] ≥ 0∧-3 + (2)y[0] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧-3 + y[0] ≥ 0 ⇒ (U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥)∧2 + (-1)Bound + (-2)y[0] + x[0] ≥ 0∧-2 + (2)y[0] ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4)    (x[0] + (-2)y[0] ≥ 0∧-3 + (2)y[0] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧-3 + y[0] ≥ 0 ⇒ (U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥)∧2 + (-1)Bound + (-2)y[0] + x[0] ≥ 0∧-2 + (2)y[0] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (x[0] + -1 + (-1)y[0] ≥ 0∧x[0] + (-2)y[0] ≥ 0∧-3 + y[0] ≥ 0∧-3 + (2)y[0] ≥ 0 ⇒ 2 + (-1)Bound + (-2)y[0] + x[0] ≥ 0∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥)∧-2 + (2)y[0] ≥ 0)

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6)    (x[0] + -4 + (-1)y[0] ≥ 0∧-6 + x[0] + (-2)y[0] ≥ 0∧y[0] ≥ 0∧3 + (2)y[0] ≥ 0 ⇒ -4 + (-1)Bound + (-2)y[0] + x[0] ≥ 0∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥)∧4 + (2)y[0] ≥ 0)

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(7)    (2 + y[0] + x[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0∧3 + (2)y[0] ≥ 0 ⇒ 2 + (-1)Bound + x[0] ≥ 0∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥)∧4 + (2)y[0] ≥ 0)

To summarize, we get the following constraints P_≥ for the following pairs.

F(TRUE, x, y) → F(&&(>@z(x, y), >@z(y, 2@z)), +@z(1@z, x), *@z(2@z, y))
- (2 + y[0] + x[0] ≥ 0∧x[0] ≥ 0∧y[0] ≥ 0∧3 + (2)y[0] ≥ 0 ⇒ 2 + (-1)Bound + x[0] ≥ 0∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >@z(y[0]1, 2@z)), +@z(1@z, x[0]1), *@z(2@z, y[0]1))), ≥)∧4 + (2)y[0] ≥ 0)

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(*@z(x₁, x₂)) = x₁·x₂
POL(TRUE) = -1
POL(&&(x₁, x₂)) = 1
POL(2@z) = 2
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = 1
POL(F(x₁, x₂, x₃)) = 1 + (-1)x₃ + x₂
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0]))

The following pairs are in P_bound:

F(TRUE, x[0], y[0]) → F(&&(>@z(x[0], y[0]), >@z(y[0], 2@z)), +@z(1@z, x[0]), *@z(2@z, y[0]))

The following pairs are in P_≥:
none

At least the following rules have been oriented under context sensitive arithmetic replacement:

*@z¹ ↔
+@z¹ ↔
&&(TRUE, TRUE)¹ → TRUE¹
FALSE¹ → &&(TRUE, FALSE)¹
&&(FALSE, TRUE)¹ → FALSE¹

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ IDP

              ↳ IDependencyGraphProof

I DP problem:
The following domains are used:none

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

f(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs.