Termination proof of /tmp/tmpimCksg/A09.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:

if(FALSE, u, v) → v
if(TRUE, u, v) → u
f(TRUE, x, y) → fNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
fNat(TRUE, x, y) → f(>@z(x, y), x, round(+@z(y, 1@z)))
round(x) → if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))

The set Q consists of the following terms:

if(FALSE, x0, x1)
if(TRUE, x0, x1)
f(TRUE, x0, x1)
fNat(TRUE, x0, x1)
round(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:

if(FALSE, u, v) → v
if(TRUE, u, v) → u
f(TRUE, x, y) → fNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)
fNat(TRUE, x, y) → f(>@z(x, y), x, round(+@z(y, 1@z)))
round(x) → if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))

The integer pair graph contains the following rules and edges:

(0): FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z)))
(1): ROUND(x[1]) → IF(=@z(%@z(x[1], 2@z), 0@z), x[1], +@z(x[1], 1@z))
(2): FNAT(TRUE, x[2], y[2]) → ROUND(+@z(y[2], 1@z))
(3): F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])

(0) -> (3), if ((x[0] →^* x[3])∧(round(+@z(y[0], 1@z)) →^* y[3])∧(>@z(x[0], y[0]) →^* TRUE))

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

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

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

The set Q consists of the following terms:

if(FALSE, x0, x1)
if(TRUE, x0, x1)
f(TRUE, x0, x1)
fNat(TRUE, x0, x1)
round(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

          ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

if(FALSE, u, v) → v
if(TRUE, u, v) → u
round(x) → if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))

(0) -> (3), if ((x[0] →^* x[3])∧(round(+@z(y[0], 1@z)) →^* y[3])∧(>@z(x[0], y[0]) →^* TRUE))

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

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

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

The set Q consists of the following terms:

if(FALSE, x0, x1)
if(TRUE, x0, x1)
f(TRUE, x0, x1)
fNat(TRUE, x0, x1)
round(x0)

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

↳ 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:

if(FALSE, u, v) → v
if(TRUE, u, v) → u
round(x) → if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))

The integer pair graph contains the following rules and edges:

(3): F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])
(0): FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z)))

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

(0) -> (3), if ((x[0] →^* x[3])∧(round(+@z(y[0], 1@z)) →^* y[3])∧(>@z(x[0], y[0]) →^* TRUE))

The set Q consists of the following terms:

if(FALSE, x0, x1)
if(TRUE, x0, x1)
f(TRUE, x0, x1)
fNat(TRUE, x0, x1)
round(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 F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3]) the following chains were created:

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

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

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

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

We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5)    (0 = 0∧0 ≥ 0∧(U^Increasing(FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])), ≥)∧0 = 0∧0 = 0∧0 ≥ 0∧0 = 0)

For Pair FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z))) the following chains were created:

We consider the chain FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z))), F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3]), FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z))), F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3]) which results in the following constraint:
(6)    (x[0]1=x[3]1∧&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z))=TRUE∧>@z(x[0]1, y[0]1)=TRUE∧x[0]=x[3]∧round(+@z(y[0]1, 1@z))=y[3]1∧x[3]=x[0]1∧>@z(x[0], y[0])=TRUE∧round(+@z(y[0], 1@z))=y[3]∧y[3]=y[0]1 ⇒ FNAT(TRUE, x[0]1, y[0]1)≥NonInfC∧FNAT(TRUE, x[0]1, y[0]1)≥F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))∧(U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥))

We simplified constraint (6) using rules (III), (IV), (VII), (IDP_BOOLEAN), (REWRITING) which results in the following new constraint:
(7)    (>@z(x[0], y[3])=TRUE∧>@z(x[0], y[0])=TRUE∧=@z(%@z(+@z(y[0], 1@z), 2@z), 0@z)=x0∧+@z(y[0], 1@z)=x1∧+@z(+@z(y[0], 1@z), 1@z)=x2∧if(x0, x1, x2)=y[3]∧>=@z(x[0], 0@z)=TRUE∧>=@z(y[3], 0@z)=TRUE ⇒ FNAT(TRUE, x[0], y[3])≥NonInfC∧FNAT(TRUE, x[0], y[3])≥F(>@z(x[0], y[3]), x[0], round(+@z(y[3], 1@z)))∧(U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥))

We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8)    (x[0] + -1 + (-1)y[3] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧(-1)x0 ≥ 0∧1 + y[0] + (-1)x1 ≥ 0∧2 + y[0] + (-1)x2 ≥ 0∧x[0] ≥ 0∧y[3] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧(-1)Bound + x[0] + (-1)y[3] ≥ 0∧0 ≥ 0)

We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9)    (x[0] + -1 + (-1)y[3] ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧(-1)x0 ≥ 0∧1 + y[0] + (-1)x1 ≥ 0∧2 + y[0] + (-1)x2 ≥ 0∧x[0] ≥ 0∧y[3] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧(-1)Bound + x[0] + (-1)y[3] ≥ 0∧0 ≥ 0)

We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10)    (1 + y[0] + (-1)x1 ≥ 0∧x[0] ≥ 0∧(-1)x0 ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧2 + y[0] + (-1)x2 ≥ 0∧y[3] ≥ 0∧x[0] + -1 + (-1)y[3] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧(-1)Bound + x[0] + (-1)y[3] ≥ 0∧0 ≥ 0)

We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(11)    (1 + y[0] + (-1)x1 ≥ 0∧x[0] ≥ 0∧x0 ≥ 0∧x[0] + -1 + (-1)y[0] ≥ 0∧2 + y[0] + (-1)x2 ≥ 0∧y[3] ≥ 0∧x[0] + -1 + (-1)y[3] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧(-1)Bound + x[0] + (-1)y[3] ≥ 0∧0 ≥ 0)

We simplified constraint (11) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(12)    (1 + y[0] + (-1)x1 ≥ 0∧1 + y[3] + x[0] ≥ 0∧x0 ≥ 0∧y[3] + x[0] + (-1)y[0] ≥ 0∧2 + y[0] + (-1)x2 ≥ 0∧y[3] ≥ 0∧x[0] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧1 + (-1)Bound + x[0] ≥ 0∧0 ≥ 0)

We simplified constraint (12) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(13)    (1 + y[3] + x[0] ≥ 0∧y[3] ≥ 0∧x[0] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧1 + (-1)Bound + x[0] ≥ 0∧0 ≥ 0)

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

F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])
- (0 = 0∧0 ≥ 0∧(U^Increasing(FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])), ≥)∧0 = 0∧0 = 0∧0 ≥ 0∧0 = 0)
FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z)))
- (1 + y[3] + x[0] ≥ 0∧y[3] ≥ 0∧x[0] ≥ 0 ⇒ (U^Increasing(F(>@z(x[0]1, y[0]1), x[0]1, round(+@z(y[0]1, 1@z)))), ≥)∧1 + (-1)Bound + x[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(0@z) = 0
POL(TRUE) = -1
POL(&&(x₁, x₂)) = -1
POL(2@z) = 2
POL(FALSE) = -1
POL(F(x₁, x₂, x₃)) = 1 + x₂ + (-1)x₃
POL(>@z(x₁, x₂)) = 0
POL(round(x₁)) = x₁
POL(=@z(x₁, x₂)) = 0
POL(if(x₁, x₂, x₃)) = (-1)max{(-1)x₂, (-1)x₃}
POL(>=@z(x₁, x₂)) = -1
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(1@z) = 1
POL(undefined) = -1
POL(FNAT(x₁, x₂, x₃)) = 2 + (2)x₁ + x₂ + (-1)x₃

The following pairs are in P_>:

F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])

The following pairs are in P_bound:

FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z)))

The following pairs are in P_≥:

FNAT(TRUE, x[0], y[0]) → F(>@z(x[0], y[0]), x[0], round(+@z(y[0], 1@z)))

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

&&(FALSE, FALSE)¹ ↔ FALSE¹
v¹ → if(FALSE, u, v)¹
u¹ → if(TRUE, u, v)¹
+@z¹ ↔
&&(TRUE, TRUE)¹ ↔ TRUE¹
&&(FALSE, TRUE)¹ ↔ FALSE¹
&&(TRUE, FALSE)¹ ↔ FALSE¹
if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))¹ → round(x)¹
%@z¹ →

↳ 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:

if(FALSE, u, v) → v
if(TRUE, u, v) → u
round(x) → if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))

The integer pair graph contains the following rules and edges:

(3): F(TRUE, x[3], y[3]) → FNAT(&&(>=@z(x[3], 0@z), >=@z(y[3], 0@z)), x[3], y[3])

The set Q consists of the following terms:

if(FALSE, x0, x1)
if(TRUE, x0, x1)
f(TRUE, x0, x1)
fNat(TRUE, x0, x1)
round(x0)

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:

if(FALSE, u, v) → v
if(TRUE, u, v) → u
round(x) → if(=@z(%@z(x, 2@z), 0@z), x, +@z(x, 1@z))

The integer pair graph contains the following rules and edges:

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

The set Q consists of the following terms:

if(FALSE, x0, x1)
if(TRUE, x0, x1)
f(TRUE, x0, x1)
fNat(TRUE, x0, x1)
round(x0)

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