Termination proof of /tmp/tmpHE3AD6/02.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:

eval(x, y) → Cond_eval(>@z(x, y), x, y)
Cond_eval(TRUE, x, y) → eval(-@z(x, 1@z), +@z(y, 1@z))

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(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:

eval(x, y) → Cond_eval(>@z(x, y), x, y)
Cond_eval(TRUE, x, y) → eval(-@z(x, 1@z), +@z(y, 1@z))

The integer pair graph contains the following rules and edges:

(0): EVAL(x[0], y[0]) → COND_EVAL(>@z(x[0], y[0]), x[0], y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))

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

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

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(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): EVAL(x[0], y[0]) → COND_EVAL(>@z(x[0], y[0]), x[0], y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))

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

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

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(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 EVAL(x, y) → COND_EVAL(>@z(x, y), x, y) the following chains were created:

We consider the chain EVAL(x[0], y[0]) → COND_EVAL(>@z(x[0], y[0]), x[0], y[0]) which results in the following constraint:
(1)    (EVAL(x[0], y[0])≥NonInfC∧EVAL(x[0], y[0])≥COND_EVAL(>@z(x[0], y[0]), x[0], y[0])∧(U^Increasing(COND_EVAL(>@z(x[0], y[0]), x[0], y[0])), ≥))

We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2)    ((U^Increasing(COND_EVAL(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3)    ((U^Increasing(COND_EVAL(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (3) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(4)    (0 ≥ 0∧(U^Increasing(COND_EVAL(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 ≥ 0)

We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5)    (0 ≥ 0∧(U^Increasing(COND_EVAL(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0)

For Pair COND_EVAL(TRUE, x, y) → EVAL(-@z(x, 1@z), +@z(y, 1@z)) the following chains were created:

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

We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
(7)    (>@z(x[0], y[0])=TRUE ⇒ COND_EVAL(TRUE, x[0], y[0])≥NonInfC∧COND_EVAL(TRUE, x[0], y[0])≥EVAL(-@z(x[0], 1@z), +@z(y[0], 1@z))∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥))

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8)    (x[0] + -1 + (-1)y[0] ≥ 0 ⇒ (U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥)∧-1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧1 ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9)    (x[0] + -1 + (-1)y[0] ≥ 0 ⇒ (U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥)∧-1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧1 ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10)    (x[0] + -1 + (-1)y[0] ≥ 0 ⇒ 1 ≥ 0∧-1 + (-1)Bound + (-1)y[0] + x[0] ≥ 0∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥))

We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(11)    (y[0] ≥ 0 ⇒ 1 ≥ 0∧(-1)Bound + y[0] ≥ 0∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥))

We simplified constraint (11) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(12)    (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 ≥ 0∧(-1)Bound + y[0] ≥ 0∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥))

(13)    (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 ≥ 0∧(-1)Bound + y[0] ≥ 0∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥))

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

EVAL(x, y) → COND_EVAL(>@z(x, y), x, y)
- (0 ≥ 0∧(U^Increasing(COND_EVAL(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0)
COND_EVAL(TRUE, x, y) → EVAL(-@z(x, 1@z), +@z(y, 1@z))
- (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 ≥ 0∧(-1)Bound + y[0] ≥ 0∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))), ≥))
- (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 ≥ 0∧(-1)Bound + y[0] ≥ 0∧(U^Increasing(EVAL(-@z(x[1], 1@z), +@z(y[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(TRUE) = -1
POL(COND_EVAL(x₁, x₂, x₃)) = -1 + (-1)x₃ + x₂
POL(EVAL(x₁, x₂)) = -1 + (-1)x₂ + x₁
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = -1
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))

The following pairs are in P_bound:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), +@z(y[1], 1@z))

The following pairs are in P_≥:

EVAL(x[0], y[0]) → COND_EVAL(>@z(x[0], y[0]), x[0], y[0])

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

-@z¹ ↔
+@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ IDP

              ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

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

(0): EVAL(x[0], y[0]) → COND_EVAL(>@z(x[0], y[0]), x[0], y[0])

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.