Termination proof of /tmp/tmpB2

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:

cd(TRUE, x) → cd(>@z(x, 0@z), -@z(x, 1@z))

The set Q consists of the following terms:

cd(TRUE, x0)

Added dependency pairs

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

cd(TRUE, x) → cd(>@z(x, 0@z), -@z(x, 1@z))

The integer pair graph contains the following rules and edges:

(0): CD(TRUE, x[0]) → CD(>@z(x[0], 0@z), -@z(x[0], 1@z))

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

The set Q consists of the following terms:

cd(TRUE, x0)

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): CD(TRUE, x[0]) → CD(>@z(x[0], 0@z), -@z(x[0], 1@z))

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

The set Q consists of the following terms:

cd(TRUE, x0)

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 CD(TRUE, x) → CD(>@z(x, 0@z), -@z(x, 1@z)) the following chains were created:

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

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

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

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

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

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

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

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

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 + (-1)x₂
POL(0@z) = 0
POL(TRUE) = 1
POL(CD(x₁, x₂)) = 2 + (2)x₂
POL(FALSE) = -1
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = 2

The following pairs are in P_>:

CD(TRUE, x[0]) → CD(>@z(x[0], 0@z), -@z(x[0], 1@z))

The following pairs are in P_bound:

CD(TRUE, x[0]) → CD(>@z(x[0], 0@z), -@z(x[0], 1@z))

The following pairs are in P_≥:
none

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

-@z¹ ↔

↳ 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:

cd(TRUE, x0)

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