Tool CaT
| Execution Time | Unknown |
|---|
| Answer | MAYBE |
|---|
| Input | SK90 2.29 |
|---|
stdout:
MAYBE
Problem:
prime(0()) -> false()
prime(s(0())) -> false()
prime(s(s(x))) -> prime1(s(s(x)),s(x))
prime1(x,0()) -> false()
prime1(x,s(0())) -> true()
prime1(x,s(s(y))) -> and(not(divp(s(s(y)),x)),prime1(x,s(y)))
divp(x,y) -> =(rem(x,y),0())
Proof:
OpenTool IRC1
| Execution Time | Unknown |
|---|
| Answer | MAYBE |
|---|
| Input | SK90 2.29 |
|---|
stdout:
MAYBE
Tool IRC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
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:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
Proof Output:
'matrix-interpretation of dimension 1' proved the best result:
Details:
--------
'matrix-interpretation of dimension 1' succeeded with the following output:
'matrix-interpretation of dimension 1'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
Proof Output:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
prime(x1) = [2] x1 + [4]
0() = [2]
false() = [1]
s(x1) = [1] x1 + [2]
prime1(x1, x2) = [0] x1 + [2] x2 + [4]
true() = [1]
and(x1, x2) = [1] x1 + [1] x2 + [1]
not(x1) = [1] x1 + [0]
divp(x1, x2) = [0] x1 + [0] x2 + [1]
=(x1, x2) = [1] x1 + [0] x2 + [0]
rem(x1, x2) = [0] x1 + [0] x2 + [0]Tool RC1
| Execution Time | Unknown |
|---|
| Answer | MAYBE |
|---|
| Input | SK90 2.29 |
|---|
stdout:
MAYBE
Tool RC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
Proof Output:
'matrix-interpretation of dimension 1' proved the best result:
Details:
--------
'matrix-interpretation of dimension 1' succeeded with the following output:
'matrix-interpretation of dimension 1'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
Proof Output:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
prime(x1) = [2] x1 + [4]
0() = [2]
false() = [1]
s(x1) = [1] x1 + [2]
prime1(x1, x2) = [0] x1 + [2] x2 + [4]
true() = [1]
and(x1, x2) = [1] x1 + [1] x2 + [1]
not(x1) = [1] x1 + [0]
divp(x1, x2) = [0] x1 + [0] x2 + [1]
=(x1, x2) = [1] x1 + [0] x2 + [0]
rem(x1, x2) = [0] x1 + [0] x2 + [0]Tool pair1rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
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:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
prime(x1) = [2 0] x1 + [0]
[0 0] [0]
0() = [2]
[0]
false() = [1]
[0]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
prime1(x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
true() = [1]
[0]
and(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
not(x1) = [1 0] x1 + [0]
[0 0] [0]
divp(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [1 1] [2]
=(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [2]
rem(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool pair2rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
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:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
prime(x1) = [2 0] x1 + [0]
[0 0] [0]
0() = [2]
[0]
false() = [1]
[0]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
prime1(x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
true() = [1]
[0]
and(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
not(x1) = [1 0] x1 + [0]
[0 0] [0]
divp(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [1 1] [2]
=(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [2]
rem(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool pair3irc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
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:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
prime(x1) = [2 0] x1 + [0]
[0 0] [0]
0() = [2]
[0]
false() = [1]
[0]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
prime1(x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
true() = [1]
[0]
and(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
not(x1) = [1 0] x1 + [0]
[0 0] [0]
divp(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [1 1] [2]
=(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [2]
rem(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool pair3rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
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:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
prime(x1) = [2 0] x1 + [0]
[0 0] [0]
0() = [2]
[0]
false() = [1]
[0]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
prime1(x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
true() = [1]
[0]
and(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
not(x1) = [1 0] x1 + [0]
[0 0] [0]
divp(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [1 1] [2]
=(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [2]
rem(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'rc (timeout of 60.0 seconds)':
----------------------------------------------
'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
prime(x1) = [2 0] x1 + [0]
[0 0] [0]
0() = [2]
[0]
false() = [1]
[0]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
prime1(x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
true() = [1]
[0]
and(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
not(x1) = [1 0] x1 + [0]
[0 0] [0]
divp(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [1 1] [2]
=(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [2]
rem(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool tup3irc
| Execution Time | 1.5654941ms |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.29 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
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:
{ prime(0()) -> false()
, prime(s(0())) -> false()
, prime(s(s(x))) -> prime1(s(s(x)), s(x))
, prime1(x, 0()) -> false()
, prime1(x, s(0())) -> true()
, prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
, divp(x, y) -> =(rem(x, y), 0())}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(prime) = {}, Uargs(s) = {}, Uargs(prime1) = {},
Uargs(and) = {1, 2}, Uargs(not) = {1}, Uargs(divp) = {},
Uargs(=) = {}, Uargs(rem) = {}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
prime(x1) = [2 0] x1 + [0]
[0 0] [0]
0() = [2]
[0]
false() = [1]
[0]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
prime1(x1, x2) = [0 0] x1 + [2 0] x2 + [0]
[0 0] [0 0] [0]
true() = [1]
[0]
and(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
not(x1) = [1 0] x1 + [0]
[0 0] [0]
divp(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [1 1] [2]
=(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [2]
rem(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
Hurray, we answered YES(?,O(n^1))