Tool CaT
stdout:
YES(?,O(n^1))
Problem:
app(nil(),xs) -> nil()
app(cons(x,xs),ys) -> cons(x,app(xs,ys))
Proof:
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {3}
transitions:
cons1(1,6) -> 5,3
cons1(2,5) -> 5,3
cons1(1,5) -> 5,3
cons1(2,6) -> 5,3
app1(1,2) -> 6*
app1(2,1) -> 5*
app1(1,1) -> 5*
app1(2,2) -> 5*
nil1() -> 6,5,3
app0(1,2) -> 3*
app0(2,1) -> 3*
app0(1,1) -> 3*
app0(2,2) -> 3*
nil0() -> 1*
cons0(1,2) -> 2*
cons0(2,1) -> 2*
cons0(1,1) -> 2*
cons0(2,2) -> 2*
problem:
QedTool IRC1
stdout:
YES(?,O(n^1))
Tool IRC2
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
Proof Output:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:
Details:
--------
'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
'Bounds with minimal-enrichment and initial automaton 'match''
--------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_1(2, 2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 3
, cons_0(2, 2) -> 2
, cons_1(2, 3) -> 1
, cons_1(2, 3) -> 3}Tool RC1
stdout:
YES(?,O(n^1))
Tool RC2
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
Proof Output:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:
Details:
--------
'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
'Bounds with minimal-enrichment and initial automaton 'match''
--------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_1(2, 2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 3
, cons_0(2, 2) -> 2
, cons_1(2, 3) -> 1
, cons_1(2, 3) -> 3}Tool pair1rc
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'pair1 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_1(2, 2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 3
, cons_0(2, 2) -> 2
, cons_1(2, 3) -> 1
, cons_1(2, 3) -> 3}
Hurray, we answered YES(?,O(n^1))Tool pair2rc
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'pair2 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with perSymbol-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_0(2, 3) -> 1
, app_0(3, 2) -> 1
, app_0(3, 3) -> 1
, app_1(2, 2) -> 4
, app_1(2, 3) -> 4
, app_1(3, 2) -> 4
, app_1(3, 3) -> 4
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, cons_0(2, 2) -> 3
, cons_0(2, 3) -> 3
, cons_0(3, 2) -> 3
, cons_0(3, 3) -> 3
, cons_1(2, 4) -> 1
, cons_1(2, 4) -> 4
, cons_1(3, 4) -> 1
, cons_1(3, 4) -> 4}
Hurray, we answered YES(?,O(n^1))Tool pair3irc
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
The input problem contains no overlaps that give rise to inapplicable rules.
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with perSymbol-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_0(2, 3) -> 1
, app_0(3, 2) -> 1
, app_0(3, 3) -> 1
, app_1(2, 2) -> 4
, app_1(2, 3) -> 4
, app_1(3, 2) -> 4
, app_1(3, 3) -> 4
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, cons_0(2, 2) -> 3
, cons_0(2, 3) -> 3
, cons_0(3, 2) -> 3
, cons_0(3, 3) -> 3
, cons_1(2, 4) -> 1
, cons_1(2, 4) -> 4
, cons_1(3, 4) -> 1
, cons_1(3, 4) -> 4}
Hurray, we answered YES(?,O(n^1))Tool pair3rc
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_1(2, 2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 3
, cons_0(2, 2) -> 2
, cons_1(2, 3) -> 1
, cons_1(2, 3) -> 3}
Hurray, we answered YES(?,O(n^1))Tool rc
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'rc (timeout of 60.0 seconds)':
----------------------------------------------
'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with perSymbol-enrichment and initial automaton 'match' (timeout of 5.0 seconds)' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_0(2, 3) -> 1
, app_0(3, 2) -> 1
, app_0(3, 3) -> 1
, app_1(2, 2) -> 4
, app_1(2, 3) -> 4
, app_1(3, 2) -> 4
, app_1(3, 3) -> 4
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, cons_0(2, 2) -> 3
, cons_0(2, 3) -> 3
, cons_0(3, 2) -> 3
, cons_0(3, 3) -> 3
, cons_1(2, 4) -> 1
, cons_1(2, 4) -> 4
, cons_1(3, 4) -> 1
, cons_1(3, 4) -> 4}
Hurray, we answered YES(?,O(n^1))Tool tup3irc
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Application of 'tup3 (timeout of 60.0 seconds)':
------------------------------------------------
The input problem contains no overlaps that give rise to inapplicable rules.
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ app(nil(), xs) -> nil()
, app(cons(x, xs), ys) -> cons(x, app(xs, ys))}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with perSymbol-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ app_0(2, 2) -> 1
, app_0(2, 3) -> 1
, app_0(3, 2) -> 1
, app_0(3, 3) -> 1
, app_1(2, 2) -> 4
, app_1(2, 3) -> 4
, app_1(3, 2) -> 4
, app_1(3, 3) -> 4
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, cons_0(2, 2) -> 3
, cons_0(2, 3) -> 3
, cons_0(3, 2) -> 3
, cons_0(3, 3) -> 3
, cons_1(2, 4) -> 1
, cons_1(2, 4) -> 4
, cons_1(3, 4) -> 1
, cons_1(3, 4) -> 4}
Hurray, we answered YES(?,O(n^1))