Termination proof of /tmp/tmpzxOf6L/indirect.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:

if1(TRUE, x, y) → h(x, y)
h(x, y) → if2(>@z(x, y), x, y)
if2(FALSE, x, y) → f(x, y)
if2(TRUE, x, y) → 0@z
if1(FALSE, x, y) → 0@z
f(x, y) → if1(>@z(x, y), x, y)

The set Q consists of the following terms:

if1(TRUE, x0, x1)
h(x0, x1)
if2(FALSE, x0, x1)
if2(TRUE, x0, x1)
if1(FALSE, 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:

if1(TRUE, x, y) → h(x, y)
h(x, y) → if2(>@z(x, y), x, y)
if2(FALSE, x, y) → f(x, y)
if2(TRUE, x, y) → 0@z
if1(FALSE, x, y) → 0@z
f(x, y) → if1(>@z(x, y), x, y)

The integer pair graph contains the following rules and edges:

(0): F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0])
(1): IF2(FALSE, x[1], y[1]) → F(x[1], y[1])
(2): IF1(TRUE, x[2], y[2]) → H(x[2], y[2])
(3): H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3])

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

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

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

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

The set Q consists of the following terms:

if1(TRUE, x0, x1)
h(x0, x1)
if2(FALSE, x0, x1)
if2(TRUE, x0, x1)
if1(FALSE, 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

          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

(0): F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0])
(1): IF2(FALSE, x[1], y[1]) → F(x[1], y[1])
(2): IF1(TRUE, x[2], y[2]) → H(x[2], y[2])
(3): H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3])

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

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

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

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

The set Q consists of the following terms:

if1(TRUE, x0, x1)
h(x0, x1)
if2(FALSE, x0, x1)
if2(TRUE, x0, x1)
if1(FALSE, 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, y) → IF1(>@z(x, y), x, y) the following chains were created:

We consider the chain F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0]) which results in the following constraint:
(1)    (F(x[0], y[0])≥NonInfC∧F(x[0], y[0])≥IF1(>@z(x[0], y[0]), x[0], y[0])∧(U^Increasing(IF1(>@z(x[0], y[0]), x[0], y[0])), ≥))

We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2)    ((U^Increasing(IF1(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3)    ((U^Increasing(IF1(>@z(x[0], y[0]), x[0], y[0])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (3) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(4)    (0 ≥ 0∧0 ≥ 0∧(U^Increasing(IF1(>@z(x[0], y[0]), x[0], y[0])), ≥))

We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5)    (0 ≥ 0∧0 = 0∧0 = 0∧0 ≥ 0∧0 = 0∧0 = 0∧(U^Increasing(IF1(>@z(x[0], y[0]), x[0], y[0])), ≥))

For Pair IF2(FALSE, x, y) → F(x, y) the following chains were created:

We consider the chain H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3]), IF2(FALSE, x[1], y[1]) → F(x[1], y[1]), F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0]) which results in the following constraint:
(6)    (x[3]=x[1]∧y[3]=y[1]∧>@z(x[3], y[3])=FALSE∧y[1]=y[0]∧x[1]=x[0] ⇒ IF2(FALSE, x[1], y[1])≥NonInfC∧IF2(FALSE, x[1], y[1])≥F(x[1], y[1])∧(U^Increasing(F(x[1], y[1])), ≥))

We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
(7)    (>@z(x[3], y[3])=FALSE ⇒ IF2(FALSE, x[3], y[3])≥NonInfC∧IF2(FALSE, x[3], y[3])≥F(x[3], y[3])∧(U^Increasing(F(x[1], y[1])), ≥))

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

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

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

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

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

(13)    (x[3] ≥ 0∧y[3] ≥ 0 ⇒ -1 + (-1)Bound + x[3] ≥ 0∧(2)x[3] ≥ 0∧(U^Increasing(F(x[1], y[1])), ≥))

For Pair IF1(TRUE, x, y) → H(x, y) the following chains were created:

We consider the chain F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0]), IF1(TRUE, x[2], y[2]) → H(x[2], y[2]), H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3]) which results in the following constraint:
(14)    (x[0]=x[2]∧y[0]=y[2]∧x[2]=x[3]∧y[2]=y[3]∧>@z(x[0], y[0])=TRUE ⇒ IF1(TRUE, x[2], y[2])≥NonInfC∧IF1(TRUE, x[2], y[2])≥H(x[2], y[2])∧(U^Increasing(H(x[2], y[2])), ≥))

We simplified constraint (14) using rules (III), (IV) which results in the following new constraint:
(15)    (>@z(x[0], y[0])=TRUE ⇒ IF1(TRUE, x[0], y[0])≥NonInfC∧IF1(TRUE, x[0], y[0])≥H(x[0], y[0])∧(U^Increasing(H(x[2], y[2])), ≥))

We simplified constraint (15) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(16)    (x[0] + -1 + (-1)y[0] ≥ 0 ⇒ (U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0∧-1 + (-2)y[0] + (2)x[0] ≥ 0)

We simplified constraint (16) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(17)    (x[0] + -1 + (-1)y[0] ≥ 0 ⇒ (U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0∧-1 + (-2)y[0] + (2)x[0] ≥ 0)

We simplified constraint (17) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(18)    (x[0] + -1 + (-1)y[0] ≥ 0 ⇒ -1 + (-2)y[0] + (2)x[0] ≥ 0∧(U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (18) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(19)    (y[0] ≥ 0 ⇒ 1 + (2)y[0] ≥ 0∧(U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (19) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(20)    (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 + (2)y[0] ≥ 0∧(U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0)

(21)    (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 + (2)y[0] ≥ 0∧(U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0)

For Pair H(x, y) → IF2(>@z(x, y), x, y) the following chains were created:

We consider the chain H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3]) which results in the following constraint:
(22)    (H(x[3], y[3])≥NonInfC∧H(x[3], y[3])≥IF2(>@z(x[3], y[3]), x[3], y[3])∧(U^Increasing(IF2(>@z(x[3], y[3]), x[3], y[3])), ≥))

We simplified constraint (22) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(23)    ((U^Increasing(IF2(>@z(x[3], y[3]), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (23) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(24)    ((U^Increasing(IF2(>@z(x[3], y[3]), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (24) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(25)    ((U^Increasing(IF2(>@z(x[3], y[3]), x[3], y[3])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (25) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(26)    (0 = 0∧(U^Increasing(IF2(>@z(x[3], y[3]), x[3], y[3])), ≥)∧0 ≥ 0∧0 = 0∧0 = 0∧0 = 0∧0 ≥ 0)

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

F(x, y) → IF1(>@z(x, y), x, y)
- (0 ≥ 0∧0 = 0∧0 = 0∧0 ≥ 0∧0 = 0∧0 = 0∧(U^Increasing(IF1(>@z(x[0], y[0]), x[0], y[0])), ≥))
IF2(FALSE, x, y) → F(x, y)
- (x[3] ≥ 0∧y[3] ≥ 0 ⇒ -1 + (-1)Bound + x[3] ≥ 0∧(2)x[3] ≥ 0∧(U^Increasing(F(x[1], y[1])), ≥))
- (x[3] ≥ 0∧y[3] ≥ 0 ⇒ -1 + (-1)Bound + x[3] ≥ 0∧(2)x[3] ≥ 0∧(U^Increasing(F(x[1], y[1])), ≥))
IF1(TRUE, x, y) → H(x, y)
- (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 + (2)y[0] ≥ 0∧(U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0)
- (y[0] ≥ 0∧x[0] ≥ 0 ⇒ 1 + (2)y[0] ≥ 0∧(U^Increasing(H(x[2], y[2])), ≥)∧0 ≥ 0)
H(x, y) → IF2(>@z(x, y), x, y)
- (0 = 0∧(U^Increasing(IF2(>@z(x[3], y[3]), x[3], y[3])), ≥)∧0 ≥ 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(IF2(x₁, x₂, x₃)) = -1 + x₃ + (-1)x₂
POL(TRUE) = -1
POL(IF1(x₁, x₂, x₃)) = -1 + (-1)x₃ + x₂
POL(FALSE) = -1
POL(F(x₁, x₂)) = -1 + (-1)x₂ + x₁
POL(undefined) = -1
POL(H(x₁, x₂)) = -1 + x₂ + (-1)x₁
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

IF1(TRUE, x[2], y[2]) → H(x[2], y[2])

The following pairs are in P_bound:

IF2(FALSE, x[1], y[1]) → F(x[1], y[1])

The following pairs are in P_≥:

F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0])
IF2(FALSE, x[1], y[1]) → F(x[1], y[1])
H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3])

There are no usable rules.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ 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:

(0): F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0])
(2): IF1(TRUE, x[2], y[2]) → H(x[2], y[2])
(3): H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3])

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

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

The set Q consists of the following terms:

if1(TRUE, x0, x1)
h(x0, x1)
if2(FALSE, x0, x1)
if2(TRUE, x0, x1)
if1(FALSE, x0, x1)
f(x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ AND

              ↳ IDP

              ↳ IDP

                ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

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

(0): F(x[0], y[0]) → IF1(>@z(x[0], y[0]), x[0], y[0])
(1): IF2(FALSE, x[1], y[1]) → F(x[1], y[1])
(3): H(x[3], y[3]) → IF2(>@z(x[3], y[3]), x[3], y[3])

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

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

The set Q consists of the following terms:

if1(TRUE, x0, x1)
h(x0, x1)
if2(FALSE, x0, x1)
if2(TRUE, x0, x1)
if1(FALSE, x0, x1)
f(x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.