Termination proof of /tmp/tmpLzU0mh/countUpExp.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:

exp(x, y) → if(>@z(y, 0@z), x, y)
cu(TRUE, x) → cu(<@z(x, exp(10@z, 2@z)), +@z(x, 1@z))
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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:

exp(x, y) → if(>@z(y, 0@z), x, y)
cu(TRUE, x) → cu(<@z(x, exp(10@z, 2@z)), +@z(x, 1@z))
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))
(1): CU(TRUE, x[1]) → EXP(10@z, 2@z)
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])
(3): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →^* x[0]a)∧(<@z(x[0], exp(10@z, 2@z)) →^* TRUE))

(0) -> (1), if ((+@z(x[0], 1@z) →^* x[1])∧(<@z(x[0], exp(10@z, 2@z)) →^* TRUE))

(1) -> (2), if (EXP(10@z, 2@z) →^* EXP(x[2], y[2]))

(2) -> (3), if ((x[2] →^* x[3])∧(y[2] →^* y[3])∧(>@z(y[2], 0@z) →^* TRUE))

(3) -> (2), if ((-@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

(0) -> (0), if ((+@z(x[0], 1@z) →^* x[0]a)∧(<@z(x[0], exp(10@z, 2@z)) →^* TRUE))

(0) -> (1), if ((+@z(x[0], 1@z) →^* x[1])∧(<@z(x[0], exp(10@z, 2@z)) →^* TRUE))

(1) -> (2), if (EXP(10@z, 2@z) →^* EXP(x[2], y[2]))

(2) -> (3), if ((x[2] →^* x[3])∧(y[2] →^* y[3])∧(>@z(y[2], 0@z) →^* TRUE))

(3) -> (2), if ((-@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ AND

              ↳ IDP

                ↳ UsableRulesProof

              ↳ IDP

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(3): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

(2) -> (3), if ((x[2] →^* x[3])∧(y[2] →^* y[3])∧(>@z(y[2], 0@z) →^* TRUE))

(3) -> (2), if ((-@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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

            ↳ AND

              ↳ IDP

                ↳ UsableRulesProof

                  ↳ IDP

                    ↳ IDPNonInfProof

              ↳ IDP

I DP problem:
The following domains are used:

z

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

(3): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

(2) -> (3), if ((x[2] →^* x[3])∧(y[2] →^* y[3])∧(>@z(y[2], 0@z) →^* TRUE))

(3) -> (2), if ((-@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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 IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z)) the following chains were created:

We consider the chain EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2]), IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z)), EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2]) which results in the following constraint:
(1)    (x[2]=x[3]∧y[2]=y[3]∧>@z(y[2], 0@z)=TRUE∧x[3]=x[2]1∧-@z(y[3], 1@z)=y[2]1 ⇒ IF(TRUE, x[3], y[3])≥NonInfC∧IF(TRUE, x[3], y[3])≥EXP(x[3], -@z(y[3], 1@z))∧(U^Increasing(EXP(x[3], -@z(y[3], 1@z))), ≥))

We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
(2)    (>@z(y[2], 0@z)=TRUE ⇒ IF(TRUE, x[2], y[2])≥NonInfC∧IF(TRUE, x[2], y[2])≥EXP(x[2], -@z(y[2], 1@z))∧(U^Increasing(EXP(x[3], -@z(y[3], 1@z))), ≥))

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

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

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

We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(6)    (y[2] + -1 ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(U^Increasing(EXP(x[3], -@z(y[3], 1@z))), ≥)∧0 = 0∧-1 + (-1)Bound + y[2] ≥ 0)

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(7)    (y[2] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(U^Increasing(EXP(x[3], -@z(y[3], 1@z))), ≥)∧0 = 0∧(-1)Bound + y[2] ≥ 0)

For Pair EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2]) the following chains were created:

We consider the chain EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2]) which results in the following constraint:
(8)    (EXP(x[2], y[2])≥NonInfC∧EXP(x[2], y[2])≥IF(>@z(y[2], 0@z), x[2], y[2])∧(U^Increasing(IF(>@z(y[2], 0@z), x[2], y[2])), ≥))

We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9)    ((U^Increasing(IF(>@z(y[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(10)    ((U^Increasing(IF(>@z(y[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (10) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(11)    (0 ≥ 0∧(U^Increasing(IF(>@z(y[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0)

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(IF(>@z(y[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0∧0 = 0)

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

IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))
- (y[2] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(U^Increasing(EXP(x[3], -@z(y[3], 1@z))), ≥)∧0 = 0∧(-1)Bound + y[2] ≥ 0)
EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])
- (0 = 0∧0 = 0∧0 = 0∧0 ≥ 0∧(U^Increasing(IF(>@z(y[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0∧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)) = x₁ + (-1)x₂
POL(0@z) = 0
POL(TRUE) = 0
POL(EXP(x₁, x₂)) = -1 + x₂
POL(FALSE) = -1
POL(IF(x₁, x₂, x₃)) = -1 + x₃
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

The following pairs are in P_bound:

IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

The following pairs are in P_≥:

EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

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

-@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ AND

              ↳ IDP

                ↳ UsableRulesProof

                  ↳ IDP

                    ↳ IDPNonInfProof

                      ↳ IDP

                        ↳ IDependencyGraphProof

              ↳ IDP

I DP problem:
The following domains are used:

z

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

(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(TRUE, x0, x1)

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ AND

              ↳ IDP

              ↳ IDP

                ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →^* x[0]a)∧(<@z(x[0], exp(10@z, 2@z)) →^* TRUE))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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 CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z)) the following chains were created:

We consider the chain CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z)), CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z)), CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@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, exp(10@z, 2@z))=TRUE∧+@z(x[0]1, 1@z)=x[0]2∧<@z(x[0], exp(10@z, 2@z))=TRUE ⇒ CU(TRUE, x[0]1)≥NonInfC∧CU(TRUE, x[0]1)≥CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥))

We simplified constraint (1) using rules (III), (IV), (IDP_CONSTANT_FOLD), (REWRITING) which results in the following new constraint:
(2)    (<@z(+@z(x[0], 1@z), 100@z)=TRUE∧<@z(x[0], 100@z)=TRUE ⇒ CU(TRUE, +@z(x[0], 1@z))≥NonInfC∧CU(TRUE, +@z(x[0], 1@z))≥CU(<@z(+@z(x[0], 1@z), exp(10@z, 2@z)), +@z(+@z(x[0], 1@z), 1@z))∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3)    (98 + (-1)x[0] ≥ 0∧99 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-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)    (98 + (-1)x[0] ≥ 0∧99 + (-1)x[0] ≥ 0 ⇒ (U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-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)    (98 + (-1)x[0] ≥ 0∧99 + (-1)x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-1)Bound + (-1)x[0] ≥ 0)

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(6)    (98 + (-1)x[0] ≥ 0∧99 + (-1)x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-1)Bound + (-1)x[0] ≥ 0)

(7)    (98 + x[0] ≥ 0∧99 + x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-1)Bound + x[0] ≥ 0)

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

CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))
- (98 + (-1)x[0] ≥ 0∧99 + (-1)x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-1)Bound + (-1)x[0] ≥ 0)
- (98 + x[0] ≥ 0∧99 + x[0] ≥ 0∧x[0] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(CU(<@z(x[0]1, exp(10@z, 2@z)), +@z(x[0]1, 1@z))), ≥)∧-2 + (-1)Bound + x[0] ≥ 0)

POL(-@z(x₁, x₂)) = x₁ + (-1)x₂
POL(exp(x₁, x₂)) = -1 + (-1)x₁
POL(0@z) = 0
POL(*@z(x₁, x₂)) = x₁·x₂
POL(TRUE) = -1
POL(2@z) = 2
POL(FALSE) = -1
POL(<@z(x₁, x₂)) = -1
POL(>@z(x₁, x₂)) = -1
POL(CU(x₁, x₂)) = -1 + (-1)x₂
POL(if(x₁, x₂, x₃)) = -1 + (-1)x₃ + (-1)x₂ + (-1)x₁
POL(10@z) = 10
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(1@z) = 1
POL(undefined) = -1

The following pairs are in P_>:

CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))

The following pairs are in P_bound:

CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@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

          ↳ IDependencyGraphProof

            ↳ AND

              ↳ IDP

              ↳ IDP

                ↳ IDPNonInfProof

                  ↳ IDP

                    ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph is empty.
The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(TRUE, x0, x1)

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