Termination proof of /tmp/tmpDz6vtp/unsatCond2.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_f(TRUE, x) → f(x)
f(x) → Cond_f(<@z(*@z(x, x), 0@z), x)

The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(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_f(TRUE, x) → f(x)
f(x) → Cond_f(<@z(*@z(x, x), 0@z), x)

The integer pair graph contains the following rules and edges:

(0): COND_F(TRUE, x[0]) → F(x[0])
(1): F(x[1]) → COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])

(0) -> (1), if ((x[0] →^* x[1]))

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

The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(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): COND_F(TRUE, x[0]) → F(x[0])
(1): F(x[1]) → COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])

(0) -> (1), if ((x[0] →^* x[1]))

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

The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(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 COND_F(TRUE, x) → F(x) the following chains were created:

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

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

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

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

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

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

(7)    (-1 + (-1)x[1]² ≥ 0∧x[1] ≥ 0 ⇒ -1 + (-1)Bound ≥ 0∧(U^Increasing(F(x[0])), ≥)∧-1 ≥ 0)

For Pair F(x) → COND_F(<@z(*@z(x, x), 0@z), x) the following chains were created:

We consider the chain F(x[1]) → COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1]) which results in the following constraint:
(8)    (F(x[1])≥NonInfC∧F(x[1])≥COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])∧(U^Increasing(COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])), ≥))

We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9)    ((U^Increasing(COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(10)    ((U^Increasing(COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (10) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(11)    (0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])), ≥))

We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(12)    (0 ≥ 0∧0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])), ≥))

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

COND_F(TRUE, x) → F(x)
- (-1 + (-1)x[1]² ≥ 0∧x[1] ≥ 0 ⇒ -1 + (-1)Bound ≥ 0∧(U^Increasing(F(x[0])), ≥)∧-1 ≥ 0)
- (-1 + (-1)x[1]² ≥ 0∧x[1] ≥ 0 ⇒ -1 + (-1)Bound ≥ 0∧(U^Increasing(F(x[0])), ≥)∧-1 ≥ 0)
F(x) → COND_F(<@z(*@z(x, x), 0@z), x)
- (0 ≥ 0∧0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])), ≥))

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(0@z) = 0
POL(*@z(x₁, x₂)) = x₁·x₂
POL(TRUE) = 2
POL(FALSE) = -1
POL(<@z(x₁, x₂)) = -1
POL(undefined) = -1
POL(F(x₁)) = -1
POL(COND_F(x₁, x₂)) = -1

The following pairs are in P_>:

COND_F(TRUE, x[0]) → F(x[0])

The following pairs are in P_bound:

COND_F(TRUE, x[0]) → F(x[0])

The following pairs are in P_≥:

F(x[1]) → COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])

There are no usable rules.

↳ 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): F(x[1]) → COND_F(<@z(*@z(x[1], x[1]), 0@z), x[1])

The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(x0)

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