Termination proof of /tmp/tmpMLK8KD/sumto_no

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_sumto(TRUE, x, y) → 0@z
sumto(x, y) → Cond_sumto1(>=@z(y, x), x, y)
Cond_sumto1(TRUE, x, y) → +@z(x, sumto(+@z(x, 1@z), y))
sumto(x, y) → Cond_sumto(>@z(x, y), x, y)

The set Q consists of the following terms:

Cond_sumto(TRUE, x0, x1)
sumto(x0, x1)
Cond_sumto1(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:

Cond_sumto(TRUE, x, y) → 0@z
sumto(x, y) → Cond_sumto1(>=@z(y, x), x, y)
Cond_sumto1(TRUE, x, y) → +@z(x, sumto(+@z(x, 1@z), y))
sumto(x, y) → Cond_sumto(>@z(x, y), x, y)

The integer pair graph contains the following rules and edges:

(0): COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0])
(1): SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])
(2): SUMTO(x[2], y[2]) → COND_SUMTO(>@z(x[2], y[2]), x[2], y[2])

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

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

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

The set Q consists of the following terms:

Cond_sumto(TRUE, x0, x1)
sumto(x0, x1)
Cond_sumto1(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): COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0])
(1): SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])
(2): SUMTO(x[2], y[2]) → COND_SUMTO(>@z(x[2], y[2]), x[2], y[2])

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

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

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

The set Q consists of the following terms:

Cond_sumto(TRUE, x0, x1)
sumto(x0, x1)
Cond_sumto1(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): SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])
(0): COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0])

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

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

The set Q consists of the following terms:

Cond_sumto(TRUE, x0, x1)
sumto(x0, x1)
Cond_sumto1(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 SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1]) the following chains were created:

We consider the chain SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1]) which results in the following constraint:
(1)    (SUMTO(x[1], y[1])≥NonInfC∧SUMTO(x[1], y[1])≥COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])∧(U^Increasing(COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])), ≥))

We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2)    ((U^Increasing(COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3)    ((U^Increasing(COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (3) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(4)    (0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])), ≥))

We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5)    (0 = 0∧(U^Increasing(COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧0 = 0)

For Pair COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0]) the following chains were created:

We consider the chain SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1]), COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0]), SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1]) which results in the following constraint:
(6)    (y[0]=y[1]1∧+@z(x[0], 1@z)=x[1]1∧>=@z(y[1], x[1])=TRUE∧y[1]=y[0]∧x[1]=x[0] ⇒ COND_SUMTO1(TRUE, x[0], y[0])≥NonInfC∧COND_SUMTO1(TRUE, x[0], y[0])≥SUMTO(+@z(x[0], 1@z), y[0])∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))

We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
(7)    (>=@z(y[1], x[1])=TRUE ⇒ COND_SUMTO1(TRUE, x[1], y[1])≥NonInfC∧COND_SUMTO1(TRUE, x[1], y[1])≥SUMTO(+@z(x[1], 1@z), y[1])∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))

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

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9)    (y[1] + (-1)x[1] ≥ 0 ⇒ (U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥)∧(-1)Bound + y[1] + (-1)x[1] ≥ 0∧0 ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10)    (y[1] + (-1)x[1] ≥ 0 ⇒ (-1)Bound + y[1] + (-1)x[1] ≥ 0∧0 ≥ 0∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))

We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(11)    (x[1] ≥ 0 ⇒ (-1)Bound + x[1] ≥ 0∧0 ≥ 0∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))

We simplified constraint (11) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(12)    (x[1] ≥ 0∧y[1] ≥ 0 ⇒ (-1)Bound + x[1] ≥ 0∧0 ≥ 0∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))

(13)    (x[1] ≥ 0∧y[1] ≥ 0 ⇒ (-1)Bound + x[1] ≥ 0∧0 ≥ 0∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))

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

SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])
- (0 = 0∧(U^Increasing(COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧0 = 0)
COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0])
- (x[1] ≥ 0∧y[1] ≥ 0 ⇒ (-1)Bound + x[1] ≥ 0∧0 ≥ 0∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[0])), ≥))
- (x[1] ≥ 0∧y[1] ≥ 0 ⇒ (-1)Bound + x[1] ≥ 0∧0 ≥ 0∧(U^Increasing(SUMTO(+@z(x[0], 1@z), y[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(SUMTO(x₁, x₂)) = x₂ + (-1)x₁
POL(>=@z(x₁, x₂)) = -1
POL(COND_SUMTO1(x₁, x₂, x₃)) = 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_>:

COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0])

The following pairs are in P_bound:

COND_SUMTO1(TRUE, x[0], y[0]) → SUMTO(+@z(x[0], 1@z), y[0])

The following pairs are in P_≥:

SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])

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:

z

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

(1): SUMTO(x[1], y[1]) → COND_SUMTO1(>=@z(y[1], x[1]), x[1], y[1])

The set Q consists of the following terms:

Cond_sumto(TRUE, x0, x1)
sumto(x0, x1)
Cond_sumto1(TRUE, x0, x1)

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