Termination proof of /tmp/tmps5nd1A/complete2.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:

Cond_eval(TRUE, x) → eval(+@z(*@z(-@z(2@z), x), 10@z))
eval(x) → Cond_eval(>=@z(x, 0@z), x)

The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(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:

Cond_eval(TRUE, x) → eval(+@z(*@z(-@z(2@z), x), 10@z))
eval(x) → Cond_eval(>=@z(x, 0@z), x)

The integer pair graph contains the following rules and edges:

(0): EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0])
(1): COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z))

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

(1) -> (0), if ((+@z(*@z(-@z(2@z), x[1]), 10@z) →^* x[0]))

The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(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): EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0])
(1): COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z))

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

(1) -> (0), if ((+@z(*@z(-@z(2@z), x[1]), 10@z) →^* x[0]))

The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(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 EVAL(x) → COND_EVAL(>=@z(x, 0@z), x) the following chains were created:

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

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

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

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

For Pair COND_EVAL(TRUE, x) → EVAL(+@z(*@z(-@z(2@z), x), 10@z)) the following chains were created:

We consider the chain COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z)), EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0]), COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z)), EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0]), COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z)), EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0]), COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z)), EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0]), COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z)), EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0]) which results in the following constraint:
(6)    (x[0]3=x[1]4∧x[0]2=x[1]3∧x[0]1=x[1]2∧>=@z(x[0]1, 0@z)=TRUE∧>=@z(x[0]2, 0@z)=TRUE∧+@z(*@z(-@z(2@z), x[1]3), 10@z)=x[0]3∧+@z(*@z(-@z(2@z), x[1]4), 10@z)=x[0]4∧x[0]=x[1]1∧+@z(*@z(-@z(2@z), x[1]1), 10@z)=x[0]1∧+@z(*@z(-@z(2@z), x[1]2), 10@z)=x[0]2∧>=@z(x[0]3, 0@z)=TRUE∧+@z(*@z(-@z(2@z), x[1]), 10@z)=x[0]∧>=@z(x[0], 0@z)=TRUE ⇒ COND_EVAL(TRUE, x[1]4)≥NonInfC∧COND_EVAL(TRUE, x[1]4)≥EVAL(+@z(*@z(-@z(2@z), x[1]4), 10@z))∧(U^Increasing(EVAL(+@z(*@z(-@z(2@z), x[1]4), 10@z))), ≥))

We simplified constraint (6) using rules (III), (IV), (IDP_CONSTANT_FOLD) which results in the following new constraint:
(7)    (>=@z(+@z(*@z(-2@z, +@z(*@z(-2@z, x[1]), 10@z)), 10@z), 0@z)=TRUE∧>=@z(+@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, x[1]), 10@z)), 10@z)), 10@z), 0@z)=TRUE∧>=@z(+@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, x[1]), 10@z)), 10@z)), 10@z)), 10@z), 0@z)=TRUE∧>=@z(+@z(*@z(-2@z, x[1]), 10@z), 0@z)=TRUE ⇒ COND_EVAL(TRUE, +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, x[1]), 10@z)), 10@z)), 10@z)), 10@z))≥NonInfC∧COND_EVAL(TRUE, +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, x[1]), 10@z)), 10@z)), 10@z)), 10@z))≥EVAL(+@z(*@z(-@z(2@z), +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, +@z(*@z(-2@z, x[1]), 10@z)), 10@z)), 10@z)), 10@z)), 10@z))∧(U^Increasing(EVAL(+@z(*@z(-@z(2@z), x[1]4), 10@z))), ≥))

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8)    ((4)x[1] + -10 ≥ 0∧(-8)x[1] + 30 ≥ 0∧(16)x[1] + -50 ≥ 0∧(-2)x[1] + 10 ≥ 0 ⇒ (U^Increasing(EVAL(+@z(*@z(-@z(2@z), x[1]4), 10@z))), ≥)∧-1 + (-1)Bound ≥ 0∧-2 ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9)    ((4)x[1] + -10 ≥ 0∧(-8)x[1] + 30 ≥ 0∧(16)x[1] + -50 ≥ 0∧(-2)x[1] + 10 ≥ 0 ⇒ (U^Increasing(EVAL(+@z(*@z(-@z(2@z), x[1]4), 10@z))), ≥)∧-1 + (-1)Bound ≥ 0∧-2 ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10)    ((4)x[1] + -10 ≥ 0∧(16)x[1] + -50 ≥ 0∧(-2)x[1] + 10 ≥ 0∧(-8)x[1] + 30 ≥ 0 ⇒ (U^Increasing(EVAL(+@z(*@z(-@z(2@z), x[1]4), 10@z))), ≥)∧-2 ≥ 0∧-1 + (-1)Bound ≥ 0)

We solved constraint (10) using rule (IDP_SMT_SPLIT).

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

EVAL(x) → COND_EVAL(>=@z(x, 0@z), x)
- ((U^Increasing(COND_EVAL(>=@z(x[0], 0@z), x[0])), ≥)∧0 ≥ 0∧0 = 0∧0 = 0∧(-1)Bound ≥ 0)
COND_EVAL(TRUE, x) → EVAL(+@z(*@z(-@z(2@z), x), 10@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)) = (-1)x₁
POL(>=@z(x₁, x₂)) = 0
POL(0@z) = 0
POL(*@z(x₁, x₂)) = x₁·x₂
POL(TRUE) = -1
POL(10@z) = 10
POL(2@z) = 2
POL(COND_EVAL(x₁, x₂)) = -1
POL(EVAL(x₁)) = 0
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = -1
POL(undefined) = -1

The following pairs are in P_>:

EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0])
COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z))

The following pairs are in P_bound:

EVAL(x[0]) → COND_EVAL(>=@z(x[0], 0@z), x[0])
COND_EVAL(TRUE, x[1]) → EVAL(+@z(*@z(-@z(2@z), x[1]), 10@z))

The following pairs are in P_≥:
none

There are no usable rules.

↳ 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:

Cond_eval(TRUE, x0)
eval(x0)

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