Tool CaT
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
Problem:
norm(nil()) -> 0()
norm(g(x,y)) -> s(norm(x))
f(x,nil()) -> g(nil(),x)
f(x,g(y,z)) -> g(f(x,y),z)
rem(nil(),y) -> nil()
rem(g(x,y),0()) -> g(x,y)
rem(g(x,y),s(z)) -> rem(x,z)
Proof:
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {7,6,5}
transitions:
rem1(3,1) -> 7*
rem1(3,3) -> 7*
rem1(4,2) -> 7*
rem1(4,4) -> 7*
rem1(1,2) -> 7*
rem1(1,4) -> 7*
rem1(2,1) -> 7*
rem1(2,3) -> 7*
rem1(3,2) -> 7*
rem1(3,4) -> 7*
rem1(4,1) -> 7*
rem1(4,3) -> 7*
rem1(1,1) -> 7*
rem1(1,3) -> 7*
rem1(2,2) -> 7*
rem1(2,4) -> 7*
g1(3,1) -> 7*
g1(3,3) -> 7*
g1(4,2) -> 7*
g1(4,4) -> 7*
g1(10,1) -> 10,6
g1(10,3) -> 10,6
g1(1,2) -> 7*
g1(1,4) -> 7*
g1(7,1) -> 10*
g1(2,1) -> 7*
g1(7,3) -> 10*
g1(2,3) -> 7*
g1(3,2) -> 7*
g1(3,4) -> 7*
g1(4,1) -> 7*
g1(4,3) -> 7*
g1(10,2) -> 10,6
g1(10,4) -> 10,6
g1(1,1) -> 7*
g1(1,3) -> 7*
g1(7,2) -> 10*
g1(2,2) -> 7*
g1(7,4) -> 10*
g1(2,4) -> 7*
nil1() -> 7,10
f1(3,1) -> 10*
f1(3,3) -> 10*
f1(4,2) -> 10*
f1(4,4) -> 10*
f1(1,2) -> 10*
f1(1,4) -> 10*
f1(2,1) -> 10*
f1(2,3) -> 10*
f1(3,2) -> 10*
f1(3,4) -> 10*
f1(4,1) -> 10*
f1(4,3) -> 10*
f1(1,1) -> 10*
f1(1,3) -> 10*
f1(2,2) -> 10*
f1(2,4) -> 10*
s1(8) -> 8,5
norm1(2) -> 8*
norm1(4) -> 8*
norm1(1) -> 8*
norm1(3) -> 8*
01() -> 8,5
norm0(2) -> 5*
norm0(4) -> 5*
norm0(1) -> 5*
norm0(3) -> 5*
nil0() -> 1*
00() -> 2*
g0(3,1) -> 3*
g0(3,3) -> 3*
g0(4,2) -> 3*
g0(4,4) -> 3*
g0(1,2) -> 3*
g0(1,4) -> 3*
g0(2,1) -> 3*
g0(2,3) -> 3*
g0(3,2) -> 3*
g0(3,4) -> 3*
g0(4,1) -> 3*
g0(4,3) -> 3*
g0(1,1) -> 3*
g0(1,3) -> 3*
g0(2,2) -> 3*
g0(2,4) -> 3*
s0(2) -> 4*
s0(4) -> 4*
s0(1) -> 4*
s0(3) -> 4*
f0(3,1) -> 6*
f0(3,3) -> 6*
f0(4,2) -> 6*
f0(4,4) -> 6*
f0(1,2) -> 6*
f0(1,4) -> 6*
f0(2,1) -> 6*
f0(2,3) -> 6*
f0(3,2) -> 6*
f0(3,4) -> 6*
f0(4,1) -> 6*
f0(4,3) -> 6*
f0(1,1) -> 6*
f0(1,3) -> 6*
f0(2,2) -> 6*
f0(2,4) -> 6*
rem0(3,1) -> 7*
rem0(3,3) -> 7*
rem0(4,2) -> 7*
rem0(4,4) -> 7*
rem0(1,2) -> 7*
rem0(1,4) -> 7*
rem0(2,1) -> 7*
rem0(2,3) -> 7*
rem0(3,2) -> 7*
rem0(3,4) -> 7*
rem0(4,1) -> 7*
rem0(4,3) -> 7*
rem0(1,1) -> 7*
rem0(1,3) -> 7*
rem0(2,2) -> 7*
rem0(2,4) -> 7*
problem:
QedTool IRC1
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
Tool IRC2
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}Tool RC1
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
Tool RC2
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}Tool pair1rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool pair2rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool pair3irc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
StartTerms: basic terms
Strategy: innermost
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:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool pair3rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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 minimal-enrichment and initial automaton 'match' (timeout of 100.0 seconds)' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool tup3irc
Execution Time | 0.10269618ms |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.41 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z)}
StartTerms: basic terms
Strategy: innermost
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:
{ norm_0(2) -> 1
, norm_1(2) -> 3
, nil_0() -> 2
, nil_1() -> 1
, nil_1() -> 4
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, g_0(2, 2) -> 2
, g_1(2, 2) -> 1
, g_1(4, 2) -> 1
, g_1(4, 2) -> 4
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, f_0(2, 2) -> 1
, f_1(2, 2) -> 4
, rem_0(2, 2) -> 1
, rem_1(2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))