interpretations
YES(?,O(n^3))
We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).
Strict Trs:
{ le(0(), Y) -> true()
, le(s(X), 0()) -> false()
, le(s(X), s(Y)) -> le(X, Y)
, minus(0(), Y) -> 0()
, minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
, ifMinus(true(), s(X), Y) -> 0()
, ifMinus(false(), s(X), Y) -> s(minus(X, Y))
, quot(0(), s(Y)) -> 0()
, quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) }
Obligation:
innermost runtime complexity
Answer:
YES(?,O(n^3))
The following argument positions are usable:
Uargs(le) = {}, Uargs(s) = {1}, Uargs(minus) = {},
Uargs(ifMinus) = {1}, Uargs(quot) = {1}
TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).
[0 1 0] [0]
[le](x1, x2) = [0 0 1] x1 + [0]
[0 0 0] [0]
[0]
[0] = [1]
[0]
[0]
[true] = [0]
[0]
[1 1 3] [1]
[s](x1) = [0 1 0] x1 + [2]
[0 1 1] [0]
[0]
[false] = [0]
[0]
[1 2 1] [3]
[minus](x1, x2) = [0 1 0] x1 + [0]
[0 0 1] [0]
[1 0 3] [1 1 1] [2]
[ifMinus](x1, x2, x3) = [0 0 0] x1 + [0 1 0] x2 + [0]
[0 0 0] [0 0 1] [0]
[2 3 3] [3 2 2] [0]
[quot](x1, x2) = [0 1 0] x1 + [0 0 0] x2 + [0]
[0 0 1] [0 0 0] [0]
This order satisfies following ordering constraints
[le(0(), Y)] = [1]
[0]
[0]
> [0]
[0]
[0]
= [true()]
[le(s(X), 0())] = [0 1 0] [2]
[0 1 1] X + [0]
[0 0 0] [0]
> [0]
[0]
[0]
= [false()]
[le(s(X), s(Y))] = [0 1 0] [2]
[0 1 1] X + [0]
[0 0 0] [0]
> [0 1 0] [0]
[0 0 1] X + [0]
[0 0 0] [0]
= [le(X, Y)]
[minus(0(), Y)] = [5]
[1]
[0]
> [0]
[1]
[0]
= [0()]
[minus(s(X), Y)] = [1 4 4] [8]
[0 1 0] X + [2]
[0 1 1] [0]
> [1 4 4] [7]
[0 1 0] X + [2]
[0 1 1] [0]
= [ifMinus(le(s(X), Y), s(X), Y)]
[ifMinus(true(), s(X), Y)] = [1 3 4] [5]
[0 1 0] X + [2]
[0 1 1] [0]
> [0]
[1]
[0]
= [0()]
[ifMinus(false(), s(X), Y)] = [1 3 4] [5]
[0 1 0] X + [2]
[0 1 1] [0]
> [1 3 4] [4]
[0 1 0] X + [2]
[0 1 1] [0]
= [s(minus(X, Y))]
[quot(0(), s(Y))] = [3 7 11] [10]
[0 0 0] Y + [1]
[0 0 0] [0]
> [0]
[1]
[0]
= [0()]
[quot(s(X), s(Y))] = [3 7 11] [2 8 9] [15]
[0 0 0] Y + [0 1 0] X + [2]
[0 0 0] [0 1 1] [0]
> [3 7 11] [2 8 8] [14]
[0 0 0] Y + [0 1 0] X + [2]
[0 0 0] [0 1 1] [0]
= [s(quot(minus(X, Y), s(Y)))]
Hurray, we answered YES(?,O(n^3))
lmpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ le(0(), Y) -> true()
, le(s(X), 0()) -> false()
, le(s(X), s(Y)) -> le(X, Y)
, minus(0(), Y) -> 0()
, minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
, ifMinus(true(), s(X), Y) -> 0()
, ifMinus(false(), s(X), Y) -> s(minus(X, Y))
, quot(0(), s(Y)) -> 0()
, quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
mpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ le(0(), Y) -> true()
, le(s(X), 0()) -> false()
, le(s(X), s(Y)) -> le(X, Y)
, minus(0(), Y) -> 0()
, minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
, ifMinus(true(), s(X), Y) -> 0()
, ifMinus(false(), s(X), Y) -> s(minus(X, Y))
, quot(0(), s(Y)) -> 0()
, quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ le(0(), Y) -> true()
, le(s(X), 0()) -> false()
, le(s(X), s(Y)) -> le(X, Y)
, minus(0(), Y) -> 0()
, minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
, ifMinus(true(), s(X), Y) -> 0()
, ifMinus(false(), s(X), Y) -> s(minus(X, Y))
, quot(0(), s(Y)) -> 0()
, quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar-ps
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ le(0(), Y) -> true()
, le(s(X), 0()) -> false()
, le(s(X), s(Y)) -> le(X, Y)
, minus(0(), Y) -> 0()
, minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
, ifMinus(true(), s(X), Y) -> 0()
, ifMinus(false(), s(X), Y) -> s(minus(X, Y))
, quot(0(), s(Y)) -> 0()
, quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y))) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..