Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → GE(x, s(y))
GE(s(u), s(v)) → GE(u, v)
The TRS R consists of the following rules:
minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → GE(x, s(y))
GE(s(u), s(v)) → GE(u, v)
The TRS R consists of the following rules:
minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GE(s(u), s(v)) → GE(u, v)
The TRS R consists of the following rules:
minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [13] we can delete all non-usable rules [14] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GE(s(u), s(v)) → GE(u, v)
R is empty.
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GE(s(u), s(v)) → GE(u, v)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- GE(s(u), s(v)) → GE(u, v)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
The TRS R consists of the following rules:
minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [13] we can delete all non-usable rules [14] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
minus(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule MINUS(x, y) → COND(ge(x, s(y)), x, y) at position [0] we obtained the following new rules:
MINUS(0, x0) → COND(false, 0, x0)
MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(0, x0) → COND(false, 0, x0)
COND(true, x, y) → MINUS(x, s(y))
MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1)
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
COND(true, x, y) → MINUS(x, s(y))
MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1)
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1) we obtained the following new rules:
MINUS(s(x0), s(z1)) → COND(ge(x0, s(z1)), s(x0), s(z1))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), s(z1)) → COND(ge(x0, s(z1)), s(x0), s(z1))
COND(true, x, y) → MINUS(x, s(y))
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND(true, x, y) → MINUS(x, s(y)) we obtained the following new rules:
COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ NonInfProof
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), s(z1)) → COND(ge(x0, s(z1)), s(x0), s(z1))
COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus with the following steps:
Note that final constraints are written in bold face.
For Pair MINUS(x, y) → COND(ge(x, s(y)), x, y) the following chains were created:
- We consider the chain MINUS(x, y) → COND(ge(x, s(y)), x, y), COND(true, x, y) → MINUS(x, s(y)) which results in the following constraint:
- (1) (COND(ge(x2, s(x3)), x2, x3)=COND(true, x4, x5) ⇒ MINUS(x2, x3)≥COND(ge(x2, s(x3)), x2, x3))
We simplified constraint (1) using rules (I), (II), (IV), (VII) which results in the following new constraint:
- (2) (s(x3)=x12∧ge(x2, x12)=true ⇒ MINUS(x2, x3)≥COND(ge(x2, s(x3)), x2, x3))
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x2, x12)=true which results in the following new constraints:
- (3) (ge(x14, x15)=true∧s(x3)=s(x15)∧(∀x16:ge(x14, x15)=true∧s(x16)=x15 ⇒ MINUS(x14, x16)≥COND(ge(x14, s(x16)), x14, x16)) ⇒ MINUS(s(x14), x3)≥COND(ge(s(x14), s(x3)), s(x14), x3))
- (4) (true=true∧s(x3)=0 ⇒ MINUS(x17, x3)≥COND(ge(x17, s(x3)), x17, x3))
We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:
- (5) (ge(x14, x15)=true ⇒ MINUS(s(x14), x15)≥COND(ge(s(x14), s(x15)), s(x14), x15))
We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on ge(x14, x15)=true which results in the following new constraints:
- (6) (ge(x19, x20)=true∧(ge(x19, x20)=true ⇒ MINUS(s(x19), x20)≥COND(ge(s(x19), s(x20)), s(x19), x20)) ⇒ MINUS(s(s(x19)), s(x20))≥COND(ge(s(s(x19)), s(s(x20))), s(s(x19)), s(x20)))
- (7) (true=true ⇒ MINUS(s(x21), 0)≥COND(ge(s(x21), s(0)), s(x21), 0))
We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (ge(x19, x20)=true ⇒ MINUS(s(x19), x20)≥COND(ge(s(x19), s(x20)), s(x19), x20)) with σ = [ ] which results in the following new constraint:
- (8) (MINUS(s(x19), x20)≥COND(ge(s(x19), s(x20)), s(x19), x20) ⇒ MINUS(s(s(x19)), s(x20))≥COND(ge(s(s(x19)), s(s(x20))), s(s(x19)), s(x20)))
We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
- (9) (MINUS(s(x21), 0)≥COND(ge(s(x21), s(0)), s(x21), 0))
For Pair COND(true, x, y) → MINUS(x, s(y)) the following chains were created:
- We consider the chain COND(true, x, y) → MINUS(x, s(y)), MINUS(x, y) → COND(ge(x, s(y)), x, y) which results in the following constraint:
- (10) (MINUS(x6, s(x7))=MINUS(x8, x9) ⇒ COND(true, x6, x7)≥MINUS(x6, s(x7)))
We simplified constraint (10) using rules (I), (II), (IV) which results in the following new constraint:
- (11) (COND(true, x6, x7)≥MINUS(x6, s(x7)))
To summarize, we get the following constraints P≥ for the following pairs.
- MINUS(x, y) → COND(ge(x, s(y)), x, y)
- (MINUS(s(x19), x20)≥COND(ge(s(x19), s(x20)), s(x19), x20) ⇒ MINUS(s(s(x19)), s(x20))≥COND(ge(s(s(x19)), s(s(x20))), s(s(x19)), s(x20)))
- (MINUS(s(x21), 0)≥COND(ge(s(x21), s(0)), s(x21), 0))
- COND(true, x, y) → MINUS(x, s(y))
- (COND(true, x6, x7)≥MINUS(x6, s(x7)))
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 [21]:
POL(0) = 0
POL(COND(x1, x2, x3)) = -1 - x1 + 2·x2 - x3
POL(MINUS(x1, x2)) = 1 + 2·x1 - x2
POL(c) = -1
POL(false) = 1
POL(ge(x1, x2)) = 0
POL(s(x1)) = 2 + x1
POL(true) = 0
The following pairs are in P>:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
The following pairs are in Pbound:
MINUS(x, y) → COND(ge(x, s(y)), x, y)
The following rules are usable:
false → ge(0, s(v))
true → ge(u, 0)
ge(u, v) → ge(s(u), s(v))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ NonInfProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
COND(true, x, y) → MINUS(x, s(y))
The TRS R consists of the following rules:
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true
The set Q consists of the following terms:
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.