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:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
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:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(if, false), f), xs) → AP(f, ap(last, xs))
AP(ap(ap(if, false), f), xs) → AP(cons, ap(f, ap(last, xs)))
AP(ap(ap(if, false), f), xs) → AP(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
AP(ap(ap(if, false), f), xs) → AP(map, f)
AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(dropLast, ap(ap(cons, y), ys))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
AP(ap(ap(if, false), f), xs) → AP(dropLast, xs)
AP(ap(ap(if, false), f), xs) → AP(last, xs)
AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(last, ap(ap(cons, y), ys))
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(if, ap(isEmpty, xs)), f)
AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
AP(ap(map, f), xs) → AP(if, ap(isEmpty, xs))
AP(ap(map, f), xs) → AP(isEmpty, xs)
The TRS R consists of the following rules:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
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:
AP(ap(ap(if, false), f), xs) → AP(f, ap(last, xs))
AP(ap(ap(if, false), f), xs) → AP(cons, ap(f, ap(last, xs)))
AP(ap(ap(if, false), f), xs) → AP(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
AP(ap(ap(if, false), f), xs) → AP(map, f)
AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(dropLast, ap(ap(cons, y), ys))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
AP(ap(ap(if, false), f), xs) → AP(dropLast, xs)
AP(ap(ap(if, false), f), xs) → AP(last, xs)
AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(last, ap(ap(cons, y), ys))
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(if, ap(isEmpty, xs)), f)
AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
AP(ap(map, f), xs) → AP(if, ap(isEmpty, xs))
AP(ap(map, f), xs) → AP(isEmpty, xs)
The TRS R consists of the following rules:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 9 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(dropLast, ap(ap(cons, y), ys))
The TRS R consists of the following rules:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
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
↳ ATransformationProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AP(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(dropLast, ap(ap(cons, y), ys))
R is empty.
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
dropLast1(cons(x, cons(y, ys))) → dropLast1(cons(y, ys))
R is empty.
The set Q consists of the following terms:
map(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
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.
map(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
dropLast1(cons(x, cons(y, ys))) → dropLast1(cons(y, ys))
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:
- dropLast1(cons(x, cons(y, ys))) → dropLast1(cons(y, ys))
The graph contains the following edges 1 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(last, ap(ap(cons, y), ys))
The TRS R consists of the following rules:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
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
↳ ATransformationProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AP(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → AP(last, ap(ap(cons, y), ys))
R is empty.
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
last1(cons(x, cons(y, ys))) → last1(cons(y, ys))
R is empty.
The set Q consists of the following terms:
map(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
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.
map(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
last1(cons(x, cons(y, ys))) → last1(cons(y, ys))
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:
- last1(cons(x, cons(y, ys))) → last1(cons(y, ys))
The graph contains the following edges 1 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(if, false), f), xs) → AP(f, ap(last, xs))
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
The TRS R consists of the following rules:
ap(ap(map, f), xs) → ap(ap(ap(if, ap(isEmpty, xs)), f), xs)
ap(ap(ap(if, true), f), xs) → nil
ap(ap(ap(if, false), f), xs) → ap(ap(cons, ap(f, ap(last, xs))), ap(ap(map, f), ap(dropLast, xs)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
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
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(if, false), f), xs) → AP(f, ap(last, xs))
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
The TRS R consists of the following rules:
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
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.
AP(ap(ap(if, false), f), xs) → AP(f, ap(last, xs))
The remaining pairs can at least be oriented weakly.
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
Used ordering: Polynomial interpretation [25]:
POL(AP(x1, x2)) = x1
POL(ap(x1, x2)) = 1 + x2
POL(cons) = 0
POL(dropLast) = 0
POL(false) = 1
POL(if) = 0
POL(isEmpty) = 0
POL(last) = 0
POL(map) = 0
POL(nil) = 1
POL(true) = 1
The following usable rules [17] were oriented:
none
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
The TRS R consists of the following rules:
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
ap(last, ap(ap(cons, x), nil)) → x
ap(last, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(last, ap(ap(cons, y), ys))
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
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
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(if, false), f), xs) → AP(ap(map, f), ap(dropLast, xs))
AP(ap(map, f), xs) → AP(ap(ap(if, ap(isEmpty, xs)), f), xs)
The TRS R consists of the following rules:
ap(dropLast, nil) → nil
ap(dropLast, ap(ap(cons, x), nil)) → nil
ap(dropLast, ap(ap(cons, x), ap(ap(cons, y), ys))) → ap(ap(cons, x), ap(dropLast, ap(ap(cons, y), ys)))
ap(isEmpty, nil) → true
ap(isEmpty, ap(ap(cons, x), xs)) → false
The set Q consists of the following terms:
ap(ap(map, x0), x1)
ap(ap(ap(if, true), x0), x1)
ap(ap(ap(if, false), x0), x1)
ap(isEmpty, nil)
ap(isEmpty, ap(ap(cons, x0), x1))
ap(last, ap(ap(cons, x0), nil))
ap(last, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
ap(dropLast, nil)
ap(dropLast, ap(ap(cons, x0), nil))
ap(dropLast, ap(ap(cons, x0), ap(ap(cons, x1), x2)))
We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
if1(false, f, xs) → map1(f, dropLast(xs))
map1(f, xs) → if1(isEmpty(xs), f, xs)
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
The set Q consists of the following terms:
map(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
isEmpty(nil)
isEmpty(cons(x0, x1))
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
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.
map(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
if1(false, f, xs) → map1(f, dropLast(xs))
map1(f, xs) → if1(isEmpty(xs), f, xs)
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
The set Q consists of the following terms:
isEmpty(nil)
isEmpty(cons(x0, x1))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule map1(f, xs) → if1(isEmpty(xs), f, xs) at position [0] we obtained the following new rules:
map1(y0, nil) → if1(true, y0, nil)
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
map1(y0, nil) → if1(true, y0, nil)
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
if1(false, f, xs) → map1(f, dropLast(xs))
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
The set Q consists of the following terms:
isEmpty(nil)
isEmpty(cons(x0, x1))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
if1(false, f, xs) → map1(f, dropLast(xs))
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
isEmpty(nil) → true
isEmpty(cons(x, xs)) → false
The set Q consists of the following terms:
isEmpty(nil)
isEmpty(cons(x0, x1))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
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
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
if1(false, f, xs) → map1(f, dropLast(xs))
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
The set Q consists of the following terms:
isEmpty(nil)
isEmpty(cons(x0, x1))
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
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.
isEmpty(nil)
isEmpty(cons(x0, x1))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
if1(false, f, xs) → map1(f, dropLast(xs))
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, nil)) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
The set Q consists of the following terms:
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
dropLast(cons(x, nil)) → nil
Used ordering: POLO with Polynomial interpretation [25]:
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(dropLast(x1)) = x1
POL(false) = 0
POL(if1(x1, x2, x3)) = 1 + 2·x1 + x2 + x3
POL(map1(x1, x2)) = 1 + x1 + x2
POL(nil) = 2
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
if1(false, f, xs) → map1(f, dropLast(xs))
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
The set Q consists of the following terms:
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
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.
map1(y0, cons(x0, x1)) → if1(false, y0, cons(x0, x1))
The remaining pairs can at least be oriented weakly.
if1(false, f, xs) → map1(f, dropLast(xs))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( cons(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( map1(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
| M( if1(x1, ..., x3) ) = | 0 | + | | · | x1 | + | | · | x2 | + | | · | x3 |
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:
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
dropLast(nil) → nil
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ ATransformationProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
if1(false, f, xs) → map1(f, dropLast(xs))
The TRS R consists of the following rules:
dropLast(nil) → nil
dropLast(cons(x, cons(y, ys))) → cons(x, dropLast(cons(y, ys)))
The set Q consists of the following terms:
dropLast(nil)
dropLast(cons(x0, nil))
dropLast(cons(x0, cons(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.