Termination proof of /tmp/tmpllUggB/a.09.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, z) → Cond_eval(&&(>=@z(x, z), >@z(y, 0@z)), x, y, z)
Cond_eval(TRUE, x, y, z) → eval(x, y, +@z(z, y))

The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)

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, z) → Cond_eval(&&(>=@z(x, z), >@z(y, 0@z)), x, y, z)
Cond_eval(TRUE, x, y, z) → eval(x, y, +@z(z, y))

The integer pair graph contains the following rules and edges:

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

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

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

The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)

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

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

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

The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)

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

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

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

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

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

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

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

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

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

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

For Pair EVAL(x, y, z) → COND_EVAL(&&(>=@z(x, z), >@z(y, 0@z)), x, y, z) the following chains were created:

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

We simplified constraint (10) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(11)    ((U^Increasing(COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (11) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(12)    ((U^Increasing(COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (12) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(13)    ((U^Increasing(COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (13) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(14)    (0 ≥ 0∧0 = 0∧(U^Increasing(COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0)

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

COND_EVAL(TRUE, x, y, z) → EVAL(x, y, +@z(z, y))
- (y[1] ≥ 0∧x[1] ≥ 0∧z[1] ≥ 0 ⇒ y[1] ≥ 0∧(U^Increasing(EVAL(x[0], y[0], +@z(z[0], y[0]))), ≥)∧(-1)Bound + x[1] ≥ 0)
- (y[1] ≥ 0∧x[1] ≥ 0∧z[1] ≥ 0 ⇒ y[1] ≥ 0∧(U^Increasing(EVAL(x[0], y[0], +@z(z[0], y[0]))), ≥)∧(-1)Bound + x[1] ≥ 0)
EVAL(x, y, z) → COND_EVAL(&&(>=@z(x, z), >@z(y, 0@z)), x, y, z)
- (0 ≥ 0∧0 = 0∧(U^Increasing(COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])), ≥)∧0 = 0∧0 = 0∧0 = 0∧0 = 0∧0 = 0∧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)) = -1
POL(0@z) = 0
POL(TRUE) = 0
POL(&&(x₁, x₂)) = 0
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(EVAL(x₁, x₂, x₃)) = (-1)x₃ + x₁
POL(COND_EVAL(x₁, x₂, x₃, x₄)) = (-1)x₄ + x₂ + (-1)x₁
POL(FALSE) = 2
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

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

The following pairs are in P_bound:

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

The following pairs are in P_≥:

EVAL(x[1], y[1], z[1]) → COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])

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

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

↳ 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:

(1): EVAL(x[1], y[1], z[1]) → COND_EVAL(&&(>=@z(x[1], z[1]), >@z(y[1], 0@z)), x[1], y[1], z[1])

The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)

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