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, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The TRS R 2 is
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The signature Sigma is {le, true, false}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → MINUS(s(x), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → MINUS(s(x), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(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 [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(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, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- LE(s(x), s(y)) → LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(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 [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(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, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- MINUS(s(x), s(y)) → MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(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 [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(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.
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y)) at position [0] we obtained the following new rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( minus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( QUOT(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.