Termination proof of /tmp/tmpxrw9Yd/extra2.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) → ft(TRUE, x, y)
ft(TRUE, x, y) → ft(>=@z(y, x), +@z(x, 1@z), y)

The set Q consists of the following terms:

f(TRUE, x0)
ft(TRUE, x0, x1)

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) → ft(TRUE, x, y)
ft(TRUE, x, y) → ft(>=@z(y, x), +@z(x, 1@z), y)

The integer pair graph contains the following rules and edges:

(0): F(TRUE, x[0]) → FT(TRUE, x[0], y[0])
(1): FT(TRUE, x[1], y[1]) → FT(>=@z(y[1], x[1]), +@z(x[1], 1@z), y[1])

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

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

The set Q consists of the following terms:

f(TRUE, x0)
ft(TRUE, x0, x1)

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

          ↳ IDependencyGraphProof

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

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

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

The set Q consists of the following terms:

f(TRUE, x0)
ft(TRUE, x0, x1)

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(1): FT(TRUE, x[1], y[1]) → FT(>=@z(y[1], x[1]), +@z(x[1], 1@z), y[1])

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

The set Q consists of the following terms:

f(TRUE, x0)
ft(TRUE, x0, x1)

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 FT(TRUE, x[1], y[1]) → FT(>=@z(y[1], x[1]), +@z(x[1], 1@z), y[1]) the following chains were created:

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

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

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

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

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

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

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

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

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

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

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

The following pairs are in P_>:

FT(TRUE, x[1], y[1]) → FT(>=@z(y[1], x[1]), +@z(x[1], 1@z), y[1])

The following pairs are in P_bound:

FT(TRUE, x[1], y[1]) → FT(>=@z(y[1], x[1]), +@z(x[1], 1@z), y[1])

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

          ↳ IDependencyGraphProof

            ↳ 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)
ft(TRUE, x0, x1)

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