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:
rand(x, y) → Cond_rand(>@z(x, 0@z), x, y)
Cond_rand(TRUE, x, y) → rand(-@z(x, 1@z), id_inc(y))
rand(x, y) → Cond_rand1(=@z(x, 0@z), x, y)
id_inc(x) → x
id_inc(x) → +@z(x, 1@z)
random(x) → Cond_random(>=@z(x, 0@z), x)
Cond_random(TRUE, x) → rand(x, 0@z)
Cond_rand1(TRUE, x, y) → y
The set Q consists of the following terms:
rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(x0)
random(x0)
Cond_random(TRUE, x0)
Cond_rand1(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:
rand(x, y) → Cond_rand(>@z(x, 0@z), x, y)
Cond_rand(TRUE, x, y) → rand(-@z(x, 1@z), id_inc(y))
rand(x, y) → Cond_rand1(=@z(x, 0@z), x, y)
id_inc(x) → x
id_inc(x) → +@z(x, 1@z)
random(x) → Cond_random(>=@z(x, 0@z), x)
Cond_random(TRUE, x) → rand(x, 0@z)
Cond_rand1(TRUE, x, y) → y
The integer pair graph contains the following rules and edges:
(0): RAND(x[0], y[0]) → COND_RAND1(=@z(x[0], 0@z), x[0], y[0])
(1): COND_RAND(TRUE, x[1], y[1]) → ID_INC(y[1])
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
(3): RANDOM(x[3]) → COND_RANDOM(>=@z(x[3], 0@z), x[3])
(4): COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
(5): COND_RANDOM(TRUE, x[5]) → RAND(x[5], 0@z)
(2) -> (1), if ((x[2] →* x[1])∧(y[2] →* y[1])∧(>@z(x[2], 0@z) →* TRUE))
(2) -> (4), if ((x[2] →* x[4])∧(y[2] →* y[4])∧(>@z(x[2], 0@z) →* TRUE))
(3) -> (5), if ((x[3] →* x[5])∧(>=@z(x[3], 0@z) →* TRUE))
(4) -> (0), if ((id_inc(y[4]) →* y[0])∧(-@z(x[4], 1@z) →* x[0]))
(4) -> (2), if ((id_inc(y[4]) →* y[2])∧(-@z(x[4], 1@z) →* x[2]))
(5) -> (0), if ((x[5] →* x[0]))
(5) -> (2), if ((x[5] →* x[2]))
The set Q consists of the following terms:
rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(x0)
random(x0)
Cond_random(TRUE, x0)
Cond_rand1(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:
id_inc(x) → x
id_inc(x) → +@z(x, 1@z)
The integer pair graph contains the following rules and edges:
(0): RAND(x[0], y[0]) → COND_RAND1(=@z(x[0], 0@z), x[0], y[0])
(1): COND_RAND(TRUE, x[1], y[1]) → ID_INC(y[1])
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
(3): RANDOM(x[3]) → COND_RANDOM(>=@z(x[3], 0@z), x[3])
(4): COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
(5): COND_RANDOM(TRUE, x[5]) → RAND(x[5], 0@z)
(2) -> (1), if ((x[2] →* x[1])∧(y[2] →* y[1])∧(>@z(x[2], 0@z) →* TRUE))
(2) -> (4), if ((x[2] →* x[4])∧(y[2] →* y[4])∧(>@z(x[2], 0@z) →* TRUE))
(3) -> (5), if ((x[3] →* x[5])∧(>=@z(x[3], 0@z) →* TRUE))
(4) -> (0), if ((id_inc(y[4]) →* y[0])∧(-@z(x[4], 1@z) →* x[0]))
(4) -> (2), if ((id_inc(y[4]) →* y[2])∧(-@z(x[4], 1@z) →* x[2]))
(5) -> (0), if ((x[5] →* x[0]))
(5) -> (2), if ((x[5] →* x[2]))
The set Q consists of the following terms:
rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(x0)
random(x0)
Cond_random(TRUE, x0)
Cond_rand1(TRUE, x0, x1)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 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:
id_inc(x) → x
id_inc(x) → +@z(x, 1@z)
The integer pair graph contains the following rules and edges:
(4): COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
(2) -> (4), if ((x[2] →* x[4])∧(y[2] →* y[4])∧(>@z(x[2], 0@z) →* TRUE))
(4) -> (2), if ((id_inc(y[4]) →* y[2])∧(-@z(x[4], 1@z) →* x[2]))
The set Q consists of the following terms:
rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(x0)
random(x0)
Cond_random(TRUE, x0)
Cond_rand1(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 COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4])) the following chains were created:
- We consider the chain RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2]), COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4])), RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2]) which results in the following constraint:
(1) (id_inc(y[4])=y[2]1∧y[2]=y[4]∧>@z(x[2], 0@z)=TRUE∧-@z(x[4], 1@z)=x[2]1∧x[2]=x[4] ⇒ COND_RAND(TRUE, x[4], y[4])≥NonInfC∧COND_RAND(TRUE, x[4], y[4])≥RAND(-@z(x[4], 1@z), id_inc(y[4]))∧(UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥))
We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
(2) (>@z(x[2], 0@z)=TRUE ⇒ COND_RAND(TRUE, x[2], y[2])≥NonInfC∧COND_RAND(TRUE, x[2], y[2])≥RAND(-@z(x[2], 1@z), id_inc(y[2]))∧(UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥))
We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3) (-1 + x[2] ≥ 0 ⇒ (UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥)∧-1 + (-1)Bound + x[2] ≥ 0∧0 ≥ 0)
We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4) (-1 + x[2] ≥ 0 ⇒ (UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥)∧-1 + (-1)Bound + x[2] ≥ 0∧0 ≥ 0)
We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5) (-1 + x[2] ≥ 0 ⇒ (UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥)∧-1 + (-1)Bound + x[2] ≥ 0∧0 ≥ 0)
We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(6) (-1 + x[2] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥)∧0 = 0∧-1 + (-1)Bound + x[2] ≥ 0)
We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(7) (x[2] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥)∧0 = 0∧(-1)Bound + x[2] ≥ 0)
For Pair RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2]) the following chains were created:
- We consider the chain RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2]) which results in the following constraint:
(8) (RAND(x[2], y[2])≥NonInfC∧RAND(x[2], y[2])≥COND_RAND(>@z(x[2], 0@z), x[2], y[2])∧(UIncreasing(COND_RAND(>@z(x[2], 0@z), x[2], y[2])), ≥))
We simplified constraint (8) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(9) ((UIncreasing(COND_RAND(>@z(x[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (9) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(10) ((UIncreasing(COND_RAND(>@z(x[2], 0@z), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (10) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(11) (0 ≥ 0∧0 ≥ 0∧(UIncreasing(COND_RAND(>@z(x[2], 0@z), x[2], y[2])), ≥))
We simplified constraint (11) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(12) (0 = 0∧0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧(UIncreasing(COND_RAND(>@z(x[2], 0@z), x[2], y[2])), ≥)∧0 = 0)
To summarize, we get the following constraints P≥ for the following pairs.
- COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
- (x[2] ≥ 0 ⇒ 0 = 0∧0 ≥ 0∧(UIncreasing(RAND(-@z(x[4], 1@z), id_inc(y[4]))), ≥)∧0 = 0∧(-1)Bound + x[2] ≥ 0)
- RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
- (0 = 0∧0 = 0∧0 ≥ 0∧0 = 0∧0 ≥ 0∧(UIncreasing(COND_RAND(>@z(x[2], 0@z), x[2], y[2])), ≥)∧0 = 0)
The constraints for P> respective Pbound 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 Pbound.
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(RAND(x1, x2)) = -1 + x1
POL(COND_RAND(x1, x2, x3)) = -1 + x2
POL(0@z) = 0
POL(TRUE) = -1
POL(id_inc(x1)) = 1 + (2)x1
POL(+@z(x1, x2)) = x1 + x2
POL(FALSE) = -1
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x1, x2)) = -1
The following pairs are in P>:
COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
The following pairs are in Pbound:
COND_RAND(TRUE, x[4], y[4]) → RAND(-@z(x[4], 1@z), id_inc(y[4]))
The following pairs are in P≥:
RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
At least the following rules have been oriented under context sensitive arithmetic replacement:
-@z1 ↔
+@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:
id_inc(x) → x
id_inc(x) → +@z(x, 1@z)
The integer pair graph contains the following rules and edges:
(2): RAND(x[2], y[2]) → COND_RAND(>@z(x[2], 0@z), x[2], y[2])
The set Q consists of the following terms:
rand(x0, x1)
Cond_rand(TRUE, x0, x1)
id_inc(x0)
random(x0)
Cond_random(TRUE, x0)
Cond_rand1(TRUE, x0, x1)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.