Tool Bounds
Execution Time | 60.02988ms |
---|
Answer | TIMEOUT |
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Input | SK90 2.38 |
---|
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ ++(++(x, y), z) -> ++(x, ++(y, z))
, ++(.(x, y), z) -> .(x, ++(y, z))
, ++(x, nil()) -> x
, ++(nil(), y) -> y}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool CDI
Execution Time | 0.15406394ms |
---|
Answer | MAYBE |
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Input | SK90 2.38 |
---|
stdout:
MAYBE
Statistics:
Number of monomials: 180
Last formula building started for bound 3
Last SAT solving started for bound 3Tool EDA
Execution Time | 0.348536ms |
---|
Answer | YES(?,O(n^2)) |
---|
Input | SK90 2.38 |
---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ ++(++(x, y), z) -> ++(x, ++(y, z))
, ++(.(x, y), z) -> .(x, ++(y, z))
, ++(x, nil()) -> x
, ++(nil(), y) -> y}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following EDA-non-satisfying matrix interpretation:
Interpretation Functions:
nil() = [1]
[0]
++(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
.(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool IDA
Execution Time | 0.48014283ms |
---|
Answer | YES(?,O(n^2)) |
---|
Input | SK90 2.38 |
---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ ++(++(x, y), z) -> ++(x, ++(y, z))
, ++(.(x, y), z) -> .(x, ++(y, z))
, ++(x, nil()) -> x
, ++(nil(), y) -> y}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following EDA-non-satisfying and IDA(2)-non-satisfying matrix interpretation:
Interpretation Functions:
nil() = [1]
[0]
++(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
.(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool TRI
Execution Time | 0.15026307ms |
---|
Answer | YES(?,O(n^2)) |
---|
Input | SK90 2.38 |
---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ ++(++(x, y), z) -> ++(x, ++(y, z))
, ++(.(x, y), z) -> .(x, ++(y, z))
, ++(x, nil()) -> x
, ++(nil(), y) -> y}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
nil() = [1]
[2]
++(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
.(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool TRI2
Execution Time | 0.11423898ms |
---|
Answer | YES(?,O(n^2)) |
---|
Input | SK90 2.38 |
---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ ++(++(x, y), z) -> ++(x, ++(y, z))
, ++(.(x, y), z) -> .(x, ++(y, z))
, ++(x, nil()) -> x
, ++(nil(), y) -> y}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
nil() = [1]
[2]
++(x1, x2) = [1 2] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
.(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
Hurray, we answered YES(?,O(n^2))