Termination proof of /tmp/tmpp3toIW/Ex09.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
minusNat(TRUE, x, y) → minus(x, round(x))
if(TRUE, u, v) → u
minus(x, y) → minusNat(&&(>=@z(y, 0@z), =@z(x, +@z(y, 1@z))), x, y)
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)
minusNat(TRUE, x0, x1)
if(TRUE, x0, x1)
minus(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
minusNat(TRUE, x, y) → minus(x, round(x))
if(TRUE, u, v) → u
minus(x, y) → minusNat(&&(>=@z(y, 0@z), =@z(x, +@z(y, 1@z))), x, y)
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): MINUS(x[0], y[0]) → MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])
(1): ROUND(x[1]) → IF(=@z(%@z(x[1], 2@z), 0@z), x[1], +@z(x[1], 1@z))
(2): MINUSNAT(TRUE, x[2], y[2]) → MINUS(x[2], round(x[2]))
(3): MINUSNAT(TRUE, x[3], y[3]) → ROUND(x[3])

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

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

(2) -> (0), if ((round(x[2]) →^* y[0])∧(x[2] →^* x[0]))

(3) -> (1), if ((x[3] →^* x[1]))

The set Q consists of the following terms:

if(FALSE, x0, x1)
minusNat(TRUE, x0, x1)
if(TRUE, x0, x1)
minus(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) -> (2), if ((x[0] →^* x[2])∧(y[0] →^* y[2])∧(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))) →^* TRUE))

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

(2) -> (0), if ((round(x[2]) →^* y[0])∧(x[2] →^* x[0]))

(3) -> (1), if ((x[3] →^* x[1]))

The set Q consists of the following terms:

if(FALSE, x0, x1)
minusNat(TRUE, x0, x1)
if(TRUE, x0, x1)
minus(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:

(2): MINUSNAT(TRUE, x[2], y[2]) → MINUS(x[2], round(x[2]))
(0): MINUS(x[0], y[0]) → MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])

(2) -> (0), if ((round(x[2]) →^* y[0])∧(x[2] →^* x[0]))

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

The set Q consists of the following terms:

if(FALSE, x0, x1)
minusNat(TRUE, x0, x1)
if(TRUE, x0, x1)
minus(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 MINUSNAT(TRUE, x[2], y[2]) → MINUS(x[2], round(x[2])) the following chains were created:

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

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

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

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4)    (-1 + (-1)x0 ≥ 0∧1 + x[2] + (-1)x1 ≥ 0∧y[0] ≥ 0∧-1 + x[2] + (-1)y[0] ≥ 0∧1 + y[0] + (-1)x[2] ≥ 0 ⇒ (U^Increasing(MINUS(x[2]1, round(x[2]1))), ≥)∧-1 + (-1)Bound + (-1)y[0] + x[2] ≥ 0∧-1 + (-1)y[0] + x[2] ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (-1 + (-1)x0 ≥ 0∧y[0] ≥ 0∧1 + x[2] + (-1)x1 ≥ 0∧-1 + x[2] + (-1)y[0] ≥ 0∧1 + y[0] + (-1)x[2] ≥ 0 ⇒ -1 + (-1)y[0] + x[2] ≥ 0∧-1 + (-1)Bound + (-1)y[0] + x[2] ≥ 0∧(U^Increasing(MINUS(x[2]1, round(x[2]1))), ≥))

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6)    (-1 + (-1)x0 ≥ 0∧y[0] ≥ 0∧2 + y[0] + x[2] + (-1)x1 ≥ 0∧x[2] ≥ 0∧(-1)x[2] ≥ 0 ⇒ x[2] ≥ 0∧(-1)Bound + x[2] ≥ 0∧(U^Increasing(MINUS(x[2]1, round(x[2]1))), ≥))

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(7)    (-1 + (-1)x0 ≥ 0∧y[0] ≥ 0∧2 + y[0] + (-1)x1 ≥ 0∧0 ≥ 0∧0 ≥ 0 ⇒ 0 ≥ 0∧(-1)Bound ≥ 0∧(U^Increasing(MINUS(x[2]1, round(x[2]1))), ≥))

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

We simplified constraint (8) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(9)    (y[0] ≥ 0∧0 ≥ 0∧0 ≥ 0 ⇒ 0 ≥ 0∧(-1)Bound ≥ 0∧(U^Increasing(MINUS(x[2]1, round(x[2]1))), ≥))

For Pair MINUS(x[0], y[0]) → MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0]) the following chains were created:

We consider the chain MINUS(x[0], y[0]) → MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0]) which results in the following constraint:
(10)    (MINUS(x[0], y[0])≥NonInfC∧MINUS(x[0], y[0])≥MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])∧(U^Increasing(MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])), ≥))

We simplified constraint (10) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(11)    ((U^Increasing(MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (11) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(12)    ((U^Increasing(MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (12) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(13)    ((U^Increasing(MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (13) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(14)    (0 = 0∧0 ≥ 0∧(U^Increasing(MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0∧0 = 0)

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

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

The following pairs are in P_>:

MINUSNAT(TRUE, x[2], y[2]) → MINUS(x[2], round(x[2]))

The following pairs are in P_bound:

MINUSNAT(TRUE, x[2], y[2]) → MINUS(x[2], round(x[2]))

The following pairs are in P_≥:

MINUS(x[0], y[0]) → MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])

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

                ↳ 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): MINUS(x[0], y[0]) → MINUSNAT(&&(>=@z(y[0], 0@z), =@z(x[0], +@z(y[0], 1@z))), x[0], y[0])

The set Q consists of the following terms:

if(FALSE, x0, x1)
minusNat(TRUE, x0, x1)
if(TRUE, x0, x1)
minus(x0, x1)
round(x0)

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