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:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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:
-1(s(x), s(y)) → -1(x, y)
GCD(s(x), s(y)) → MIN(x, y)
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → -1(max(x, y), min(x, y))
GCD(s(x), s(y)) → MAX(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
GCD(s(x), s(y)) → MIN(x, y)
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → -1(max(x, y), min(x, y))
GCD(s(x), s(y)) → MAX(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 3 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(x, y)
R is empty.
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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.
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(s(x0), s(x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
-1(s(x), s(y)) → -1(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:
- -1(s(x), s(y)) → -1(x, y)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(x, y)
R is empty.
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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.
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(s(x0), s(x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MAX(s(x), s(y)) → MAX(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:
- MAX(s(x), s(y)) → MAX(x, y)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(x, y)
R is empty.
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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.
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(s(x0), s(x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MIN(s(x), s(y)) → MIN(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:
- MIN(s(x), s(y)) → MIN(x, y)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
The TRS R consists of the following rules:
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), 0) → s(x)
gcd(0, s(x)) → s(x)
gcd(s(x), s(y)) → gcd(-(max(x, y), min(x, y)), s(min(x, y)))
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(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.
gcd(s(x0), 0)
gcd(0, s(x0))
gcd(s(x0), s(x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(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.
GCD(s(x), s(y)) → GCD(-(max(x, y), min(x, y)), s(min(x, y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( -(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( max(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( min(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( GCD(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:
max(x, 0) → x
max(0, y) → y
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
The set Q consists of the following terms:
min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(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.