Termination proof of /tmp/tmpd0jJxK/countup.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:

cu(TRUE, x) → cu(<@z(x, 100000@z), +@z(x, 1@z))

The set Q consists of the following terms:

cu(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:

cu(TRUE, x) → cu(<@z(x, 100000@z), +@z(x, 1@z))

The integer pair graph contains the following rules and edges:

(0): CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →^* x[0]a)∧(<@z(x[0], 100000@z) →^* TRUE))

The set Q consists of the following terms:

cu(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): CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →^* x[0]a)∧(<@z(x[0], 100000@z) →^* TRUE))

The set Q consists of the following terms:

cu(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 CU(TRUE, x) → CU(<@z(x, 100000@z), +@z(x, 1@z)) the following chains were created:

We consider the chain CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z)), CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z)), CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z)) which results in the following constraint:
(1)    (+@z(x[0], 1@z)=x[0]1∧<@z(x[0]1, 100000@z)=TRUE∧+@z(x[0]1, 1@z)=x[0]2∧<@z(x[0], 100000@z)=TRUE ⇒ CU(TRUE, x[0]1)≥NonInfC∧CU(TRUE, x[0]1)≥CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥))

We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
(2)    (<@z(+@z(x[0], 1@z), 100000@z)=TRUE∧<@z(x[0], 100000@z)=TRUE ⇒ CU(TRUE, +@z(x[0], 1@z))≥NonInfC∧CU(TRUE, +@z(x[0], 1@z))≥CU(<@z(+@z(x[0], 1@z), 100000@z), +@z(+@z(x[0], 1@z), 1@z))∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3)    (99998 + (-1)x[0] ≥ 0∧99999 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + (-1)x[0] ≥ 0∧0 ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4)    (99998 + (-1)x[0] ≥ 0∧99999 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + (-1)x[0] ≥ 0∧0 ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (99998 + (-1)x[0] ≥ 0∧99999 + (-1)x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + (-1)x[0] ≥ 0)

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(6)    (99998 + (-1)x[0] ≥ 0∧99999 + (-1)x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + (-1)x[0] ≥ 0)

(7)    (99998 + x[0] ≥ 0∧99999 + x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + x[0] ≥ 0)

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

CU(TRUE, x) → CU(<@z(x, 100000@z), +@z(x, 1@z))
- (99998 + (-1)x[0] ≥ 0∧99999 + (-1)x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + (-1)x[0] ≥ 0)
- (99998 + x[0] ≥ 0∧99999 + x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, 100000@z), +@z(x[0]1, 1@z))), ≥)∧-1 + (-1)Bound + x[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(CU(x₁, x₂)) = (-1)x₂
POL(TRUE) = -1
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = -1
POL(100000@z) = 100000
POL(<@z(x₁, x₂)) = -1
POL(1@z) = 1
POL(undefined) = -1

The following pairs are in P_>:

CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z))

The following pairs are in P_bound:

CU(TRUE, x[0]) → CU(<@z(x[0], 100000@z), +@z(x[0], 1@z))

The following pairs are in P_≥:
none

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

+@z¹ ↔

↳ 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:

cu(TRUE, x0)

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