Termination proof of /tmp/tmpL0372h/extra1.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:

f(TRUE, x) → f(&&(>@z(x, y), >=@z(x, 0@z)), y)

The set Q consists of the following terms:

f(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:

f(TRUE, x) → f(&&(>@z(x, y), >=@z(x, 0@z)), y)

The integer pair graph contains the following rules and edges:

(0): F(TRUE, x[0]) → F(&&(>@z(x[0], y[0]), >=@z(x[0], 0@z)), y[0])

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

The set Q consists of the following terms:

f(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): F(TRUE, x[0]) → F(&&(>@z(x[0], y[0]), >=@z(x[0], 0@z)), y[0])

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

The set Q consists of the following terms:

f(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 F(TRUE, x) → F(&&(>@z(x, y), >=@z(x, 0@z)), y) the following chains were created:

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

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

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

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

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

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

(7)    (y[0] ≥ 0∧x[0] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧-1 + y[0] + y[0]1 ≥ 0∧y[0]1 ≥ 0 ⇒ -1 + y[0] + y[0]1 ≥ 0∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >=@z(x[0]1, 0@z)), y[0]1)), ≥)∧(-1)Bound + y[0] ≥ 0)

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(8)    (1 + y[0]1 + y[0] ≥ 0∧x[0] ≥ 0∧x[0] + -2 + (-1)y[0]1 + (-1)y[0] ≥ 0∧y[0] ≥ 0∧y[0]1 ≥ 0 ⇒ y[0] ≥ 0∧(U^Increasing(F(&&(>@z(x[0]1, y[0]1), >=@z(x[0]1, 0@z)), y[0]1)), ≥)∧1 + (-1)Bound + y[0]1 + y[0] ≥ 0)

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

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

The following pairs are in P_>:

F(TRUE, x[0]) → F(&&(>@z(x[0], y[0]), >=@z(x[0], 0@z)), y[0])

The following pairs are in P_bound:

F(TRUE, x[0]) → F(&&(>@z(x[0], y[0]), >=@z(x[0], 0@z)), y[0])

The following pairs are in P_≥:
none

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

&&(FALSE, FALSE)¹ ↔ FALSE¹
&&(TRUE, TRUE)¹ ↔ TRUE¹
&&(TRUE, FALSE)¹ ↔ FALSE¹
&&(FALSE, TRUE)¹ ↔ FALSE¹

↳ 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:

f(TRUE, x0)

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