Termination proof of /tmp/tmpSCMaJJ/A10.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
fNat(TRUE, x, y) → f(>@z(x, y), trunc(x), +@z(y, 1@z))
trunc(x) → if(=@z(%@z(x, 2@z), 0@z), x, -@z(x, 1@z))
if(TRUE, u, v) → u
f(TRUE, x, y) → fNat(&&(>=@z(x, 0@z), >=@z(y, 0@z)), x, y)

The set Q consists of the following terms:

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

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

The integer pair graph contains the following rules and edges:

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

(0) -> (3), if ((x[0] →^* x[3]))

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

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

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

The set Q consists of the following terms:

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

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

(0) -> (3), if ((x[0] →^* x[3]))

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

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

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

The set Q consists of the following terms:

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

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
trunc(x) → if(=@z(%@z(x, 2@z), 0@z), x, -@z(x, 1@z))
if(TRUE, u, v) → u

The integer pair graph contains the following rules and edges:

(1): FNAT(TRUE, x[1], y[1]) → F(>@z(x[1], y[1]), trunc(x[1]), +@z(y[1], 1@z))
(2): F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2])

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

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

The set Q consists of the following terms:

if(FALSE, x0, x1)
fNat(TRUE, x0, x1)
trunc(x0)
if(TRUE, x0, x1)
f(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 FNAT(TRUE, x[1], y[1]) → F(>@z(x[1], y[1]), trunc(x[1]), +@z(y[1], 1@z)) the following chains were created:

We consider the chain FNAT(TRUE, x[1], y[1]) → F(>@z(x[1], y[1]), trunc(x[1]), +@z(y[1], 1@z)), F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2]), FNAT(TRUE, x[1], y[1]) → F(>@z(x[1], y[1]), trunc(x[1]), +@z(y[1], 1@z)), F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2]) which results in the following constraint:
(1)    (>@z(x[1], y[1])=TRUE∧+@z(y[1], 1@z)=y[2]∧trunc(x[1]1)=x[2]1∧+@z(y[1]1, 1@z)=y[2]1∧x[2]=x[1]1∧>@z(x[1]1, y[1]1)=TRUE∧trunc(x[1])=x[2]∧y[2]=y[1]1∧&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z))=TRUE ⇒ FNAT(TRUE, x[1]1, y[1]1)≥NonInfC∧FNAT(TRUE, x[1]1, y[1]1)≥F(>@z(x[1]1, y[1]1), trunc(x[1]1), +@z(y[1]1, 1@z))∧(U^Increasing(F(>@z(x[1]1, y[1]1), trunc(x[1]1), +@z(y[1]1, 1@z))), ≥))

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

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

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

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

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

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

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

For Pair F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2]) the following chains were created:

We consider the chain F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2]) which results in the following constraint:
(9)    (F(TRUE, x[2], y[2])≥NonInfC∧F(TRUE, x[2], y[2])≥FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2])∧(U^Increasing(FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2])), ≥))

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

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

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

We simplified constraint (12) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(13)    (0 = 0∧0 = 0∧0 = 0∧(U^Increasing(FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0∧0 = 0)

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

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

The following pairs are in P_>:

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

The following pairs are in P_bound:

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

The following pairs are in P_≥:

F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2])

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

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

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ IDP

                  ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

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

The integer pair graph contains the following rules and edges:

(2): F(TRUE, x[2], y[2]) → FNAT(&&(>=@z(x[2], 0@z), >=@z(y[2], 0@z)), x[2], y[2])

The set Q consists of the following terms:

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

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