Termination proof of /tmp/tmpSHl4Lm/sqrt.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:

Cond_f1(TRUE, x, z, u, w) → z
f(x, z, u, w) → Cond_f1(>@z(w, x), x, z, u, w)
f(x, z, u, w) → Cond_f(&&(>=@z(x, w), >=@z(u, 0@z)), x, z, u, w)
sqrt(x) → f(x, 0@z, 1@z, 1@z)
Cond_f(TRUE, x, z, u, w) → f(x, +@z(z, 1@z), +@z(u, 2@z), +@z(+@z(w, u), 2@z))

The set Q consists of the following terms:

Cond_f1(TRUE, x0, x1, x2, x3)
f(x0, x1, x2, x3)
sqrt(x0)
Cond_f(TRUE, x0, x1, x2, x3)

Added dependency pairs

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

Cond_f1(TRUE, x, z, u, w) → z
f(x, z, u, w) → Cond_f1(>@z(w, x), x, z, u, w)
f(x, z, u, w) → Cond_f(&&(>=@z(x, w), >=@z(u, 0@z)), x, z, u, w)
sqrt(x) → f(x, 0@z, 1@z, 1@z)
Cond_f(TRUE, x, z, u, w) → f(x, +@z(z, 1@z), +@z(u, 2@z), +@z(+@z(w, u), 2@z))

The integer pair graph contains the following rules and edges:

(0): F(x[0], z[0], u[0], w[0]) → COND_F1(>@z(w[0], x[0]), x[0], z[0], u[0], w[0])
(1): F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])
(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))
(3): SQRT(x[3]) → F(x[3], 0@z, 1@z, 1@z)

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

(2) -> (0), if ((+@z(+@z(w[2], u[2]), 2@z) →^* w[0])∧(+@z(z[2], 1@z) →^* z[0])∧(+@z(u[2], 2@z) →^* u[0])∧(x[2] →^* x[0]))

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

(3) -> (0), if ((x[3] →^* x[0]))

(3) -> (1), if ((x[3] →^* x[1]))

The set Q consists of the following terms:

Cond_f1(TRUE, x0, x1, x2, x3)
f(x0, x1, x2, x3)
sqrt(x0)
Cond_f(TRUE, x0, x1, x2, x3)

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

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

(0): F(x[0], z[0], u[0], w[0]) → COND_F1(>@z(w[0], x[0]), x[0], z[0], u[0], w[0])
(1): F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])
(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))
(3): SQRT(x[3]) → F(x[3], 0@z, 1@z, 1@z)

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

(2) -> (0), if ((+@z(+@z(w[2], u[2]), 2@z) →^* w[0])∧(+@z(z[2], 1@z) →^* z[0])∧(+@z(u[2], 2@z) →^* u[0])∧(x[2] →^* x[0]))

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

(3) -> (0), if ((x[3] →^* x[0]))

(3) -> (1), if ((x[3] →^* x[1]))

The set Q consists of the following terms:

Cond_f1(TRUE, x0, x1, x2, x3)
f(x0, x1, x2, x3)
sqrt(x0)
Cond_f(TRUE, x0, x1, x2, x3)

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

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

(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))
(1): F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])

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

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

The set Q consists of the following terms:

Cond_f1(TRUE, x0, x1, x2, x3)
f(x0, x1, x2, x3)
sqrt(x0)
Cond_f(TRUE, x0, x1, x2, x3)

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 COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z)) the following chains were created:

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

We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
(2)    (>=@z(x[1], w[1])=TRUE∧>=@z(u[1], 0@z)=TRUE ⇒ COND_F(TRUE, x[1], z[1], u[1], w[1])≥NonInfC∧COND_F(TRUE, x[1], z[1], u[1], w[1])≥F(x[1], +@z(z[1], 1@z), +@z(u[1], 2@z), +@z(+@z(w[1], u[1]), 2@z))∧(U^Increasing(F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))), ≥))

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

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

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

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

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

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

(9)    (u[1] ≥ 0∧x[1] ≥ 0∧w[1] ≥ 0 ⇒ 0 = 0∧-1 + (-1)Bound + x[1] ≥ 0∧1 + u[1] ≥ 0∧(U^Increasing(F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))), ≥)∧0 = 0)

For Pair F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1]) the following chains were created:

We consider the chain F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1]) which results in the following constraint:
(10)    (F(x[1], z[1], u[1], w[1])≥NonInfC∧F(x[1], z[1], u[1], w[1])≥COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])∧(U^Increasing(COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])), ≥))

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

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

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

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

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

COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))
- (u[1] ≥ 0∧x[1] ≥ 0∧w[1] ≥ 0 ⇒ 0 = 0∧-1 + (-1)Bound + x[1] ≥ 0∧1 + u[1] ≥ 0∧(U^Increasing(F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))), ≥)∧0 = 0)
- (u[1] ≥ 0∧x[1] ≥ 0∧w[1] ≥ 0 ⇒ 0 = 0∧-1 + (-1)Bound + x[1] ≥ 0∧1 + u[1] ≥ 0∧(U^Increasing(F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))), ≥)∧0 = 0)
F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])
- (0 = 0∧0 = 0∧0 ≥ 0∧0 ≥ 0∧0 = 0∧(U^Increasing(COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])), ≥)∧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(F(x₁, x₂, x₃, x₄)) = -1 + (-1)x₄ + x₁
POL(>=@z(x₁, x₂)) = -1
POL(COND_F(x₁, x₂, x₃, x₄, x₅)) = -1 + (-1)x₅ + x₂
POL(0@z) = 0
POL(TRUE) = 0
POL(&&(x₁, x₂)) = -1
POL(2@z) = 2
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = -1
POL(1@z) = 1
POL(undefined) = -1

The following pairs are in P_>:

COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))

The following pairs are in P_bound:

COND_F(TRUE, x[2], z[2], u[2], w[2]) → F(x[2], +@z(z[2], 1@z), +@z(u[2], 2@z), +@z(+@z(w[2], u[2]), 2@z))

The following pairs are in P_≥:

F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])

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

&&(FALSE, FALSE)¹ ↔ FALSE¹
TRUE¹ → &&(TRUE, TRUE)¹
+@z¹ ↔
&&(FALSE, TRUE)¹ ↔ FALSE¹
FALSE¹ → &&(TRUE, FALSE)¹

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ IDP

                  ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

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

(1): F(x[1], z[1], u[1], w[1]) → COND_F(&&(>=@z(x[1], w[1]), >=@z(u[1], 0@z)), x[1], z[1], u[1], w[1])

The set Q consists of the following terms:

Cond_f1(TRUE, x0, x1, x2, x3)
f(x0, x1, x2, x3)
sqrt(x0)
Cond_f(TRUE, x0, x1, x2, x3)

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