Termination proof of /tmp/tmpC

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:

round(x) → +@z(x, 1@z)
Cond_f(TRUE, x, y) → f(x, round(x))
round(x) → x
f(x, y) → Cond_f(&&(>=@z(x, 1@z), =@z(y, -@z(x, 1@z))), x, y)

The set Q consists of the following terms:

round(x0)
Cond_f(TRUE, x0, x1)
f(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:

round(x) → +@z(x, 1@z)
Cond_f(TRUE, x, y) → f(x, round(x))
round(x) → x
f(x, y) → Cond_f(&&(>=@z(x, 1@z), =@z(y, -@z(x, 1@z))), x, y)

The integer pair graph contains the following rules and edges:

(0): COND_F(TRUE, x[0], y[0]) → F(x[0], round(x[0]))
(1): F(x[1], y[1]) → COND_F(&&(>=@z(x[1], 1@z), =@z(y[1], -@z(x[1], 1@z))), x[1], y[1])
(2): COND_F(TRUE, x[2], y[2]) → ROUND(x[2])

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

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

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

The set Q consists of the following terms:

round(x0)
Cond_f(TRUE, x0, x1)
f(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:

round(x) → +@z(x, 1@z)
round(x) → x

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

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

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

The set Q consists of the following terms:

round(x0)
Cond_f(TRUE, x0, x1)
f(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

The ITRS R consists of the following rules:

round(x) → +@z(x, 1@z)
round(x) → x

The integer pair graph contains the following rules and edges:

(1): F(x[1], y[1]) → COND_F(&&(>=@z(x[1], 1@z), =@z(y[1], -@z(x[1], 1@z))), x[1], y[1])
(0): COND_F(TRUE, x[0], y[0]) → F(x[0], round(x[0]))

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

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

The set Q consists of the following terms:

round(x0)
Cond_f(TRUE, x0, x1)
f(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 F(x[1], y[1]) → COND_F(&&(>=@z(x[1], 1@z), =@z(y[1], -@z(x[1], 1@z))), x[1], y[1]) the following chains were created:

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

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

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

For Pair COND_F(TRUE, x[0], y[0]) → F(x[0], round(x[0])) the following chains were created:

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

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

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

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

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

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

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

We simplified constraint (12) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(13)    (0 ≥ 0∧0 ≥ 0∧y[1] ≥ 0 ⇒ 1 + (-1)Bound ≥ 0∧0 ≥ 0∧(U^Increasing(F(x[0], round(x[0]))), ≥))

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

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

The following pairs are in P_>:

F(x[1], y[1]) → COND_F(&&(>=@z(x[1], 1@z), =@z(y[1], -@z(x[1], 1@z))), x[1], y[1])

The following pairs are in P_bound:

COND_F(TRUE, x[0], y[0]) → F(x[0], round(x[0]))

The following pairs are in P_≥:

COND_F(TRUE, x[0], y[0]) → F(x[0], round(x[0]))

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

&&(FALSE, FALSE)¹ ↔ FALSE¹
+@z(x, 1@z)¹ → round(x)¹
x¹ → round(x)¹
+@z¹ ↔
&&(TRUE, TRUE)¹ ↔ TRUE¹
&&(FALSE, TRUE)¹ ↔ FALSE¹
&&(TRUE, FALSE)¹ ↔ FALSE¹

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ AND

                  ↳ IDP

                    ↳ IDependencyGraphProof

                  ↳ IDP

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

round(x) → +@z(x, 1@z)
round(x) → x

The integer pair graph contains the following rules and edges:

(0): COND_F(TRUE, x[0], y[0]) → F(x[0], round(x[0]))

The set Q consists of the following terms:

round(x0)
Cond_f(TRUE, x0, x1)
f(x0, x1)

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ AND

                  ↳ IDP

                  ↳ IDP

                    ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

round(x) → +@z(x, 1@z)
round(x) → x

The integer pair graph contains the following rules and edges:

(1): F(x[1], y[1]) → COND_F(&&(>=@z(x[1], 1@z), =@z(y[1], -@z(x[1], 1@z))), x[1], y[1])

The set Q consists of the following terms:

round(x0)
Cond_f(TRUE, x0, x1)
f(x0, x1)

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