Termination proof of /tmp/tmpmJcjmu/A14.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:

cond2(FALSE, x, y) → +@z(1@z, diff(+@z(x, 1@z), y))
cond1(TRUE, x, y) → 0@z
cond2(TRUE, x, y) → +@z(1@z, diff(x, +@z(y, 1@z)))
diff(x, y) → cond1(=@z(x, y), x, y)
cond1(FALSE, x, y) → cond2(>@z(x, y), x, y)

The set Q consists of the following terms:

cond2(FALSE, x0, x1)
cond1(TRUE, x0, x1)
cond2(TRUE, x0, x1)
diff(x0, x1)
cond1(FALSE, 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:

cond2(FALSE, x, y) → +@z(1@z, diff(+@z(x, 1@z), y))
cond1(TRUE, x, y) → 0@z
cond2(TRUE, x, y) → +@z(1@z, diff(x, +@z(y, 1@z)))
diff(x, y) → cond1(=@z(x, y), x, y)
cond1(FALSE, x, y) → cond2(>@z(x, y), x, y)

The integer pair graph contains the following rules and edges:

(0): COND2(FALSE, x[0], y[0]) → DIFF(+@z(x[0], 1@z), y[0])
(1): COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1])
(2): DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
(3): COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

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

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

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

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

(3) -> (2), if ((+@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

cond2(FALSE, x0, x1)
cond1(TRUE, x0, x1)
cond2(TRUE, x0, x1)
diff(x0, x1)
cond1(FALSE, 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): COND2(FALSE, x[0], y[0]) → DIFF(+@z(x[0], 1@z), y[0])
(1): COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1])
(2): DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
(3): COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

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

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

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

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

(3) -> (2), if ((+@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

cond2(FALSE, x0, x1)
cond1(TRUE, x0, x1)
cond2(TRUE, x0, x1)
diff(x0, x1)
cond1(FALSE, 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 COND2(FALSE, x, y) → DIFF(+@z(x, 1@z), y) the following chains were created:

We consider the chain DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2]), COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1]), COND2(FALSE, x[0], y[0]) → DIFF(+@z(x[0], 1@z), y[0]) which results in the following constraint:
(1)    (y[2]=y[1]∧x[2]=x[1]∧=@z(x[2], y[2])=FALSE∧y[1]=y[0]∧x[1]=x[0]∧>@z(x[1], y[1])=FALSE ⇒ COND2(FALSE, x[0], y[0])≥NonInfC∧COND2(FALSE, x[0], y[0])≥DIFF(+@z(x[0], 1@z), y[0])∧(U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥))

We simplified constraint (1) using rules (III), (IDP_BOOLEAN) which results in the following new constraint:
(2)    (>@z(x[2], y[2])=FALSE∧<@z(x[2], y[2])=TRUE ⇒ COND2(FALSE, x[2], y[2])≥NonInfC∧COND2(FALSE, x[2], y[2])≥DIFF(+@z(x[2], 1@z), y[2])∧(U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥))

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

We simplified constraint (4) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(5)    (y[2] + (-1)x[2] ≥ 0∧-1 + x[2] + (-1)y[2] ≥ 0 ⇒ (U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥)∧-1 + (-1)Bound + max{(-1)x[2] + y[2], x[2] + (-1)y[2]} ≥ 0∧-1 + max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{-1 + (-1)x[2] + y[2], 1 + x[2] + (-1)y[2]} ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(6)    (y[2] + (-1)x[2] ≥ 0∧y[2] + -1 + (-1)x[2] ≥ 0 ⇒ (U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥)∧-1 + (-1)Bound + max{(-1)x[2] + y[2], x[2] + (-1)y[2]} ≥ 0∧-1 + max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{-1 + (-1)x[2] + y[2], 1 + x[2] + (-1)y[2]} ≥ 0)

We solved constraint (5) using rule (POLY_REMOVE_MIN_MAX).We simplified constraint (6) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(7)    ((-2)x[2] + (2)y[2] ≥ 0∧y[2] + -1 + (-1)x[2] ≥ 0∧-2 + (-2)x[2] + (2)y[2] ≥ 0∧y[2] + (-1)x[2] ≥ 0 ⇒ -1 + (-1)Bound + (-1)x[2] + y[2] ≥ 0∧0 ≥ 0∧(U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥))

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

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

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

(11)    (2 + (2)x[2] ≥ 0∧x[2] ≥ 0∧(2)x[2] ≥ 0∧1 + x[2] ≥ 0∧y[2] ≥ 0 ⇒ (-1)Bound + x[2] ≥ 0∧0 ≥ 0∧(U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥))

For Pair COND1(FALSE, x, y) → COND2(>@z(x, y), x, y) the following chains were created:

We consider the chain COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1]) which results in the following constraint:
(12)    (COND1(FALSE, x[1], y[1])≥NonInfC∧COND1(FALSE, x[1], y[1])≥COND2(>@z(x[1], y[1]), x[1], y[1])∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (12) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(13)    ((U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (13) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(14)    ((U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (14) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraints:
(15)    ((-2)x[1] + (2)y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

(16)    (-1 + (2)x[1] + (-2)y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (15) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(17)    ((2)x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (16) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(18)    (-1 + (2)x[1] + (-2)y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

(19)    (-1 + (-2)x[1] + (-2)y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (17) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(20)    ((2)x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

(21)    ((2)x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (18) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(22)    (-1 + (2)x[1] + (-2)y[1] ≥ 0∧x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

(23)    (-1 + (2)x[1] + (2)y[1] ≥ 0∧x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (19) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(24)    (-1 + (-2)x[1] + (2)y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

For Pair DIFF(x, y) → COND1(=@z(x, y), x, y) the following chains were created:

We consider the chain DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2]) which results in the following constraint:
(25)    (DIFF(x[2], y[2])≥NonInfC∧DIFF(x[2], y[2])≥COND1(=@z(x[2], y[2]), x[2], y[2])∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥))

We simplified constraint (25) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(26)    ((U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0∧max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{(-1)x[2] + y[2], x[2] + (-1)y[2]} ≥ 0)

We simplified constraint (26) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(27)    ((U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0∧max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{(-1)x[2] + y[2], x[2] + (-1)y[2]} ≥ 0)

We simplified constraint (27) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraints:
(28)    ((-2)x[2] + (2)y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

(29)    (-1 + (2)x[2] + (-2)y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (28) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(30)    ((2)x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (29) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(31)    (-1 + (2)x[2] + (-2)y[2] ≥ 0∧x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

(32)    (-1 + (-2)x[2] + (-2)y[2] ≥ 0∧x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (30) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(33)    ((2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

(34)    ((2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (31) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(35)    (-1 + (2)x[2] + (-2)y[2] ≥ 0∧x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

(36)    (-1 + (2)x[2] + (2)y[2] ≥ 0∧x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (32) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(37)    (-1 + (-2)x[2] + (2)y[2] ≥ 0∧x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

For Pair COND2(TRUE, x, y) → DIFF(x, +@z(y, 1@z)) the following chains were created:

We consider the chain DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2]), COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1]), COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z)) which results in the following constraint:
(38)    (y[2]=y[1]∧>@z(x[1], y[1])=TRUE∧y[1]=y[3]∧x[2]=x[1]∧=@z(x[2], y[2])=FALSE∧x[1]=x[3] ⇒ COND2(TRUE, x[3], y[3])≥NonInfC∧COND2(TRUE, x[3], y[3])≥DIFF(x[3], +@z(y[3], 1@z))∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥))

We simplified constraint (38) using rules (III), (IDP_BOOLEAN) which results in the following new constraint:
(39)    (>@z(x[2], y[2])=TRUE∧<@z(x[2], y[2])=TRUE ⇒ COND2(TRUE, x[2], y[2])≥NonInfC∧COND2(TRUE, x[2], y[2])≥DIFF(x[2], +@z(y[2], 1@z))∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥))

We simplified constraint (39) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(40)    (-1 + x[2] + (-1)y[2] ≥ 0∧y[2] + -1 + (-1)x[2] ≥ 0 ⇒ (U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧0 ≥ 0∧max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{1 + (-1)x[2] + y[2], -1 + x[2] + (-1)y[2]} ≥ 0)

We simplified constraint (41) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(42)    (-1 + x[2] + (-1)y[2] ≥ 0 ⇒ (U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧0 ≥ 0∧max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{1 + (-1)x[2] + y[2], -1 + x[2] + (-1)y[2]} ≥ 0)

We simplified constraint (40) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(43)    (-1 + x[2] + (-1)y[2] ≥ 0∧y[2] + -1 + (-1)x[2] ≥ 0 ⇒ (U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧0 ≥ 0∧max{(-1)x[2] + y[2], x[2] + (-1)y[2]} + (-1)max{1 + (-1)x[2] + y[2], -1 + x[2] + (-1)y[2]} ≥ 0)

We simplified constraint (42) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraints:
(44)    (2 + (-2)x[2] + (2)y[2] ≥ 0∧-1 + (2)x[2] + (-2)y[2] ≥ 0∧-1 + x[2] + (-1)y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧-1 + (2)x[2] + (-2)y[2] ≥ 0)

(45)    (-1 + (2)x[2] + (-2)y[2] ≥ 0∧-1 + x[2] + (-1)y[2] ≥ 0∧-3 + (2)x[2] + (-2)y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

We solved constraint (43) using rule (POLY_REMOVE_MIN_MAX).We simplified constraint (44) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(46)    ((-2)x[2] ≥ 0∧1 + (2)x[2] ≥ 0∧x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 + (2)x[2] ≥ 0)

We simplified constraint (45) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(47)    (1 + (2)x[2] ≥ 0∧x[2] ≥ 0∧-1 + (2)x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

We simplified constraint (46) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(48)    (0 ≥ 0∧1 ≥ 0∧0 ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

We simplified constraint (48) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(49)    (0 ≥ 0∧1 ≥ 0∧0 ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

(50)    (0 ≥ 0∧1 ≥ 0∧0 ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

We simplified constraint (47) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(51)    (1 + (2)x[2] ≥ 0∧x[2] ≥ 0∧-1 + (2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

(52)    (1 + (2)x[2] ≥ 0∧x[2] ≥ 0∧-1 + (2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)

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

COND2(FALSE, x, y) → DIFF(+@z(x, 1@z), y)
- (2 + (2)x[2] ≥ 0∧x[2] ≥ 0∧(2)x[2] ≥ 0∧1 + x[2] ≥ 0∧y[2] ≥ 0 ⇒ (-1)Bound + x[2] ≥ 0∧0 ≥ 0∧(U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥))
- (2 + (2)x[2] ≥ 0∧x[2] ≥ 0∧(2)x[2] ≥ 0∧1 + x[2] ≥ 0∧y[2] ≥ 0 ⇒ (-1)Bound + x[2] ≥ 0∧0 ≥ 0∧(U^Increasing(DIFF(+@z(x[0], 1@z), y[0])), ≥))
COND1(FALSE, x, y) → COND2(>@z(x, y), x, y)
- ((2)x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))
- ((2)x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))
- (-1 + (2)x[1] + (-2)y[1] ≥ 0∧x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))
- (-1 + (2)x[1] + (2)y[1] ≥ 0∧x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))
- (-1 + (-2)x[1] + (2)y[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧0 ≥ 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))
DIFF(x, y) → COND1(=@z(x, y), x, y)
- ((2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)
- ((2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)
- (-1 + (2)x[2] + (-2)y[2] ≥ 0∧x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)
- (-1 + (2)x[2] + (2)y[2] ≥ 0∧x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)
- (-1 + (-2)x[2] + (2)y[2] ≥ 0∧x[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)
COND2(TRUE, x, y) → DIFF(x, +@z(y, 1@z))
- (0 ≥ 0∧1 ≥ 0∧0 ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)
- (0 ≥ 0∧1 ≥ 0∧0 ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)
- (1 + (2)x[2] ≥ 0∧x[2] ≥ 0∧-1 + (2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 0)
- (1 + (2)x[2] ≥ 0∧x[2] ≥ 0∧-1 + (2)x[2] ≥ 0∧y[2] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧1 ≥ 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(DIFF(x₁, x₂)) = -1 + max{(-1)x₁ + x₂, x₁ + (-1)x₂}
POL(=@z(x₁, x₂)) = -1
POL(TRUE) = -1
POL(COND2(x₁, x₂, x₃)) = -1 + max{(-1)x₂ + x₃, x₂ + (-1)x₃}
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = -1
POL(COND1(x₁, x₂, x₃)) = -1 + max{(-1)x₂ + x₃, x₂ + (-1)x₃}
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

COND2(FALSE, x[0], y[0]) → DIFF(+@z(x[0], 1@z), y[0])

The following pairs are in P_bound:

COND2(FALSE, x[0], y[0]) → DIFF(+@z(x[0], 1@z), y[0])

The following pairs are in P_≥:

COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1])
DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

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

+@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ IDP

              ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

(1): COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1])
(2): DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
(3): COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

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

(3) -> (2), if ((+@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

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

The set Q consists of the following terms:

cond2(FALSE, x0, x1)
cond1(TRUE, x0, x1)
cond2(TRUE, x0, x1)
diff(x0, x1)
cond1(FALSE, 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 COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1]) the following chains were created:

We consider the chain COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1]) which results in the following constraint:
(1)    (COND1(FALSE, x[1], y[1])≥NonInfC∧COND1(FALSE, x[1], y[1])≥COND2(>@z(x[1], y[1]), x[1], y[1])∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2)    ((U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3)    ((U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥)∧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(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

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(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))

For Pair DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2]) the following chains were created:

We consider the chain DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2]) which results in the following constraint:
(6)    (DIFF(x[2], y[2])≥NonInfC∧DIFF(x[2], y[2])≥COND1(=@z(x[2], y[2]), x[2], y[2])∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥))

We simplified constraint (6) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(7)    ((U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (7) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(8)    ((U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (8) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(9)    (0 ≥ 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (9) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(10)    (0 ≥ 0∧0 = 0∧0 = 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0)

For Pair COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z)) the following chains were created:

We consider the chain COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1]), COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z)), DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2]) which results in the following constraint:
(11)    (>@z(x[1], y[1])=TRUE∧y[1]=y[3]∧x[1]=x[3]∧x[3]=x[2]∧+@z(y[3], 1@z)=y[2] ⇒ COND2(TRUE, x[3], y[3])≥NonInfC∧COND2(TRUE, x[3], y[3])≥DIFF(x[3], +@z(y[3], 1@z))∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥))

We simplified constraint (11) using rules (III), (IV) which results in the following new constraint:
(12)    (>@z(x[1], y[1])=TRUE ⇒ COND2(TRUE, x[1], y[1])≥NonInfC∧COND2(TRUE, x[1], y[1])≥DIFF(x[1], +@z(y[1], 1@z))∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥))

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

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

We simplified constraint (14) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(15)    (x[1] + -1 + (-1)y[1] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧-1 + (-1)Bound + (-1)y[1] + x[1] ≥ 0)

We simplified constraint (15) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(16)    (x[1] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧(-1)Bound + x[1] ≥ 0)

We simplified constraint (16) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(17)    (x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧(-1)Bound + x[1] ≥ 0)

(18)    (x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧(-1)Bound + x[1] ≥ 0)

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

COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1])
- (0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧0 = 0∧0 = 0∧(U^Increasing(COND2(>@z(x[1], y[1]), x[1], y[1])), ≥))
DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
- (0 ≥ 0∧0 = 0∧0 = 0∧(U^Increasing(COND1(=@z(x[2], y[2]), x[2], y[2])), ≥)∧0 = 0∧0 ≥ 0∧0 = 0)
COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))
- (x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧(-1)Bound + x[1] ≥ 0)
- (x[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧(U^Increasing(DIFF(x[3], +@z(y[3], 1@z))), ≥)∧(-1)Bound + x[1] ≥ 0)

POL(DIFF(x₁, x₂)) = (-1)x₂ + x₁
POL(=@z(x₁, x₂)) = 1
POL(TRUE) = -1
POL(COND2(x₁, x₂, x₃)) = -1 + (-1)x₃ + x₂
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(FALSE) = 2
POL(1@z) = 1
POL(COND1(x₁, x₂, x₃)) = (-1)x₃ + x₂
POL(undefined) = -1
POL(>@z(x₁, x₂)) = -1

The following pairs are in P_>:

COND1(FALSE, x[1], y[1]) → COND2(>@z(x[1], y[1]), x[1], y[1])

The following pairs are in P_bound:

COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

The following pairs are in P_≥:

DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

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

+@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ AND

                  ↳ IDP

                    ↳ IDependencyGraphProof

                  ↳ IDP

I DP problem:
The following domains are used:

z

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

The set Q consists of the following terms:

cond2(FALSE, x0, x1)
cond1(TRUE, x0, x1)
cond2(TRUE, x0, x1)
diff(x0, x1)
cond1(FALSE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDPNonInfProof

            ↳ 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:

(2): DIFF(x[2], y[2]) → COND1(=@z(x[2], y[2]), x[2], y[2])
(3): COND2(TRUE, x[3], y[3]) → DIFF(x[3], +@z(y[3], 1@z))

(3) -> (2), if ((+@z(y[3], 1@z) →^* y[2])∧(x[3] →^* x[2]))

The set Q consists of the following terms:

cond2(FALSE, x0, x1)
cond1(TRUE, x0, x1)
cond2(TRUE, x0, x1)
diff(x0, x1)
cond1(FALSE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.