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:
lastbit(x) → Cond_lastbit(>@z(x, 1@z), x)
conv(0@z) → cons(nil, 0@z)
conv(x) → Cond_conv(>@z(x, 0@z), x)
lastbit(1@z) → 1@z
Cond_conv(TRUE, x) → cons(conv(/@z(x, 2@z)), lastbit(x))
Cond_lastbit(TRUE, x) → lastbit(-@z(x, 2@z))
lastbit(0@z) → 0@z
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, 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:
lastbit(x) → Cond_lastbit(>@z(x, 1@z), x)
conv(0@z) → cons(nil, 0@z)
conv(x) → Cond_conv(>@z(x, 0@z), x)
lastbit(1@z) → 1@z
Cond_conv(TRUE, x) → cons(conv(/@z(x, 2@z)), lastbit(x))
Cond_lastbit(TRUE, x) → lastbit(-@z(x, 2@z))
lastbit(0@z) → 0@z
The integer pair graph contains the following rules and edges:
(0): COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z))
(1): CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1])
(2): COND_CONV(TRUE, x[2]) → LASTBIT(x[2])
(3): LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3])
(4): COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z))
(0) -> (3), if ((-@z(x[0], 2@z) →* x[3]))
(1) -> (2), if ((x[1] →* x[2])∧(>@z(x[1], 0@z) →* TRUE))
(1) -> (4), if ((x[1] →* x[4])∧(>@z(x[1], 0@z) →* TRUE))
(2) -> (3), if ((x[2] →* x[3]))
(3) -> (0), if ((x[3] →* x[0])∧(>@z(x[3], 1@z) →* TRUE))
(4) -> (1), if ((/@z(x[4], 2@z) →* x[1]))
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, 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
R is empty.
The integer pair graph contains the following rules and edges:
(0): COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z))
(1): CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1])
(2): COND_CONV(TRUE, x[2]) → LASTBIT(x[2])
(3): LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3])
(4): COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z))
(0) -> (3), if ((-@z(x[0], 2@z) →* x[3]))
(1) -> (2), if ((x[1] →* x[2])∧(>@z(x[1], 0@z) →* TRUE))
(1) -> (4), if ((x[1] →* x[4])∧(>@z(x[1], 0@z) →* TRUE))
(2) -> (3), if ((x[2] →* x[3]))
(3) -> (0), if ((x[3] →* x[0])∧(>@z(x[3], 1@z) →* TRUE))
(4) -> (1), if ((/@z(x[4], 2@z) →* x[1]))
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, x0)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
↳ ITRS
↳ ITRStoIDPProof
↳ IDP
↳ UsableRulesProof
↳ IDP
↳ IDependencyGraphProof
↳ AND
↳ IDP
↳ IDPNonInfProof
↳ IDP
I DP problem:
The following domains are used:
z
R is empty.
The integer pair graph contains the following rules and edges:
(3): LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3])
(0): COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z))
(0) -> (3), if ((-@z(x[0], 2@z) →* x[3]))
(3) -> (0), if ((x[3] →* x[0])∧(>@z(x[3], 1@z) →* TRUE))
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, 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 LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3]) the following chains were created:
- We consider the chain LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3]) which results in the following constraint:
(1) (LASTBIT(x[3])≥NonInfC∧LASTBIT(x[3])≥COND_LASTBIT(>@z(x[3], 1@z), x[3])∧(UIncreasing(COND_LASTBIT(>@z(x[3], 1@z), x[3])), ≥))
We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2) ((UIncreasing(COND_LASTBIT(>@z(x[3], 1@z), x[3])), ≥)∧0 ≥ 0∧1 ≥ 0)
We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3) ((UIncreasing(COND_LASTBIT(>@z(x[3], 1@z), x[3])), ≥)∧0 ≥ 0∧1 ≥ 0)
We simplified constraint (3) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(4) ((UIncreasing(COND_LASTBIT(>@z(x[3], 1@z), x[3])), ≥)∧0 ≥ 0∧1 ≥ 0)
We simplified constraint (4) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(5) (0 = 0∧(UIncreasing(COND_LASTBIT(>@z(x[3], 1@z), x[3])), ≥)∧1 ≥ 0∧0 = 0∧0 ≥ 0)
For Pair COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z)) the following chains were created:
- We consider the chain LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3]), COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z)), LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3]) which results in the following constraint:
(6) (>@z(x[3], 1@z)=TRUE∧x[3]=x[0]∧-@z(x[0], 2@z)=x[3]1 ⇒ COND_LASTBIT(TRUE, x[0])≥NonInfC∧COND_LASTBIT(TRUE, x[0])≥LASTBIT(-@z(x[0], 2@z))∧(UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥))
We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
(7) (>@z(x[3], 1@z)=TRUE ⇒ COND_LASTBIT(TRUE, x[3])≥NonInfC∧COND_LASTBIT(TRUE, x[3])≥LASTBIT(-@z(x[3], 2@z))∧(UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥))
We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8) (-2 + x[3] ≥ 0 ⇒ (UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥)∧(-1)Bound + (2)x[3] ≥ 0∧1 ≥ 0)
We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9) (-2 + x[3] ≥ 0 ⇒ (UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥)∧(-1)Bound + (2)x[3] ≥ 0∧1 ≥ 0)
We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10) (-2 + x[3] ≥ 0 ⇒ (-1)Bound + (2)x[3] ≥ 0∧(UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥)∧1 ≥ 0)
We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(11) (x[3] ≥ 0 ⇒ 4 + (-1)Bound + (2)x[3] ≥ 0∧(UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥)∧1 ≥ 0)
To summarize, we get the following constraints P≥ for the following pairs.
- LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3])
- (0 = 0∧(UIncreasing(COND_LASTBIT(>@z(x[3], 1@z), x[3])), ≥)∧1 ≥ 0∧0 = 0∧0 ≥ 0)
- COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z))
- (x[3] ≥ 0 ⇒ 4 + (-1)Bound + (2)x[3] ≥ 0∧(UIncreasing(LASTBIT(-@z(x[0], 2@z))), ≥)∧1 ≥ 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(TRUE) = 0
POL(2@z) = 2
POL(LASTBIT(x1)) = 2 + (2)x1
POL(FALSE) = -1
POL(COND_LASTBIT(x1, x2)) = (2)x2
POL(1@z) = 1
POL(undefined) = -1
POL(>@z(x1, x2)) = 2
The following pairs are in P>:
LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3])
COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z))
The following pairs are in Pbound:
COND_LASTBIT(TRUE, x[0]) → LASTBIT(-@z(x[0], 2@z))
The following pairs are in P≥:
none
At least the following rules have been oriented under context sensitive arithmetic replacement:
-@z1 ↔
↳ ITRS
↳ ITRStoIDPProof
↳ IDP
↳ UsableRulesProof
↳ IDP
↳ IDependencyGraphProof
↳ AND
↳ IDP
↳ IDPNonInfProof
↳ AND
↳ IDP
↳ IDependencyGraphProof
↳ IDP
↳ IDP
I DP problem:
The following domains are used:
z
R is empty.
The integer pair graph contains the following rules and edges:
(3): LASTBIT(x[3]) → COND_LASTBIT(>@z(x[3], 1@z), x[3])
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, x0)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
↳ ITRS
↳ ITRStoIDPProof
↳ IDP
↳ UsableRulesProof
↳ IDP
↳ IDependencyGraphProof
↳ AND
↳ IDP
↳ IDPNonInfProof
↳ AND
↳ IDP
↳ IDP
↳ IDependencyGraphProof
↳ IDP
I DP problem:
The following domains are used:none
R is empty.
The integer pair graph is empty.
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, x0)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs.
↳ ITRS
↳ ITRStoIDPProof
↳ IDP
↳ UsableRulesProof
↳ IDP
↳ IDependencyGraphProof
↳ AND
↳ IDP
↳ 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): CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1])
(4): COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z))
(1) -> (4), if ((x[1] →* x[4])∧(>@z(x[1], 0@z) →* TRUE))
(4) -> (1), if ((/@z(x[4], 2@z) →* x[1]))
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, 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 CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1]) the following chains were created:
- We consider the chain CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1]) which results in the following constraint:
(1) (CONV(x[1])≥NonInfC∧CONV(x[1])≥COND_CONV(>@z(x[1], 0@z), x[1])∧(UIncreasing(COND_CONV(>@z(x[1], 0@z), x[1])), ≥))
We simplified constraint (1) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(2) ((UIncreasing(COND_CONV(>@z(x[1], 0@z), x[1])), ≥)∧0 ≥ 0∧0 ≥ 0)
We simplified constraint (2) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(3) ((UIncreasing(COND_CONV(>@z(x[1], 0@z), x[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∧(UIncreasing(COND_CONV(>@z(x[1], 0@z), x[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∧(UIncreasing(COND_CONV(>@z(x[1], 0@z), x[1])), ≥))
For Pair COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z)) the following chains were created:
- We consider the chain CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1]), COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z)), CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1]) which results in the following constraint:
(6) (>@z(x[1], 0@z)=TRUE∧/@z(x[4], 2@z)=x[1]1∧x[1]=x[4] ⇒ COND_CONV(TRUE, x[4])≥NonInfC∧COND_CONV(TRUE, x[4])≥CONV(/@z(x[4], 2@z))∧(UIncreasing(CONV(/@z(x[4], 2@z))), ≥))
We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
(7) (>@z(x[1], 0@z)=TRUE ⇒ COND_CONV(TRUE, x[1])≥NonInfC∧COND_CONV(TRUE, x[1])≥CONV(/@z(x[1], 2@z))∧(UIncreasing(CONV(/@z(x[4], 2@z))), ≥))
We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8) (x[1] + -1 ≥ 0 ⇒ (UIncreasing(CONV(/@z(x[4], 2@z))), ≥)∧-1 + (-1)Bound + (2)x[1] ≥ 0∧(2)x[1] + (-2)max{x[1], (-1)x[1]} ≥ 0)
We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9) (x[1] + -1 ≥ 0 ⇒ (UIncreasing(CONV(/@z(x[4], 2@z))), ≥)∧-1 + (-1)Bound + (2)x[1] ≥ 0∧(2)x[1] + (-2)max{x[1], (-1)x[1]} ≥ 0)
We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10) ((2)x[1] ≥ 0∧x[1] + -1 ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(CONV(/@z(x[4], 2@z))), ≥)∧-1 + (-1)Bound + (2)x[1] ≥ 0)
We simplified constraint (10) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(11) (2 + (2)x[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(CONV(/@z(x[4], 2@z))), ≥)∧1 + (-1)Bound + (2)x[1] ≥ 0)
To summarize, we get the following constraints P≥ for the following pairs.
- CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1])
- (0 = 0∧0 ≥ 0∧0 ≥ 0∧0 = 0∧(UIncreasing(COND_CONV(>@z(x[1], 0@z), x[1])), ≥))
- COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z))
- (2 + (2)x[1] ≥ 0∧x[1] ≥ 0 ⇒ 0 ≥ 0∧(UIncreasing(CONV(/@z(x[4], 2@z))), ≥)∧1 + (-1)Bound + (2)x[1] ≥ 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(COND_CONV(x1, x2)) = -1 + (2)x2
POL(0@z) = 0
POL(CONV(x1)) = (2)x1
POL(TRUE) = 1
POL(2@z) = 2
POL(FALSE) = -1
POL(undefined) = -1
POL(>@z(x1, x2)) = -1
Polynomial Interpretations with Context Sensitive Arithemetic Replacement
POL(TermCSAR-Mode @ Context)
POL(/@z(x1, 2@z)1 @ {CONV_1/0}) = -1 + max{x1, (-1)x1}
The following pairs are in P>:
CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1])
COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z))
The following pairs are in Pbound:
COND_CONV(TRUE, x[4]) → CONV(/@z(x[4], 2@z))
The following pairs are in P≥:
none
At least the following rules have been oriented under context sensitive arithmetic replacement:
/@z1 →
↳ ITRS
↳ ITRStoIDPProof
↳ IDP
↳ UsableRulesProof
↳ IDP
↳ IDependencyGraphProof
↳ AND
↳ IDP
↳ IDP
↳ IDPNonInfProof
↳ AND
↳ IDP
↳ IDependencyGraphProof
↳ IDP
I DP problem:
The following domains are used:none
R is empty.
The integer pair graph is empty.
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, x0)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs.
↳ ITRS
↳ ITRStoIDPProof
↳ IDP
↳ UsableRulesProof
↳ IDP
↳ IDependencyGraphProof
↳ AND
↳ IDP
↳ 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:
(1): CONV(x[1]) → COND_CONV(>@z(x[1], 0@z), x[1])
The set Q consists of the following terms:
lastbit(x0)
conv(x0)
Cond_conv(TRUE, x0)
Cond_lastbit(TRUE, x0)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.