We are left with following problem, upon which TcT provides the
certificate YES(?,POLY).
Strict Trs:
{ f(s(x1), x2, x3, x4, x5, x6, x7, x8, x9, x10) ->
f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10)
, f(0(), s(x2), x3, x4, x5, x6, x7, x8, x9, x10) ->
f(x2, x2, x3, x4, x5, x6, x7, x8, x9, x10)
, f(0(), 0(), s(x3), x4, x5, x6, x7, x8, x9, x10) ->
f(x3, x3, x3, x4, x5, x6, x7, x8, x9, x10)
, f(0(), 0(), 0(), s(x4), x5, x6, x7, x8, x9, x10) ->
f(x4, x4, x4, x4, x5, x6, x7, x8, x9, x10)
, f(0(), 0(), 0(), 0(), s(x5), x6, x7, x8, x9, x10) ->
f(x5, x5, x5, x5, x5, x6, x7, x8, x9, x10)
, f(0(), 0(), 0(), 0(), 0(), s(x6), x7, x8, x9, x10) ->
f(x6, x6, x6, x6, x6, x6, x7, x8, x9, x10)
, f(0(), 0(), 0(), 0(), 0(), 0(), s(x7), x8, x9, x10) ->
f(x7, x7, x7, x7, x7, x7, x7, x8, x9, x10)
, f(0(), 0(), 0(), 0(), 0(), 0(), 0(), s(x8), x9, x10) ->
f(x8, x8, x8, x8, x8, x8, x8, x8, x9, x10)
, f(0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), s(x9), x10) ->
f(x9, x9, x9, x9, x9, x9, x9, x9, x9, x10)
, f(0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), s(x10)) ->
f(x10, x10, x10, x10, x10, x10, x10, x10, x10, x10)
, f(0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), 0()) -> 0() }
Obligation:
innermost runtime complexity
Answer:
YES(?,POLY)
The input was oriented with the instance of 'Polynomial Path Order
(PS)' as induced by the safe mapping
safe(f) = {}, safe(s) = {1}, safe(0) = {}
and precedence
empty .
Following symbols are considered recursive:
{f}
The recursion depth is 1.
For your convenience, here are the satisfied ordering constraints:
f(s(; x1), x2, x3, x4, x5, x6, x7, x8, x9, x10;) > f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10;)
f(0(), s(; x2), x3, x4, x5, x6, x7, x8, x9, x10;) > f(x2, x2, x3, x4, x5, x6, x7, x8, x9, x10;)
f(0(), 0(), s(; x3), x4, x5, x6, x7, x8, x9, x10;) > f(x3, x3, x3, x4, x5, x6, x7, x8, x9, x10;)
f(0(), 0(), 0(), s(; x4), x5, x6, x7, x8, x9, x10;) > f(x4, x4, x4, x4, x5, x6, x7, x8, x9, x10;)
f(0(), 0(), 0(), 0(), s(; x5), x6, x7, x8, x9, x10;) > f(x5, x5, x5, x5, x5, x6, x7, x8, x9, x10;)
f(0(), 0(), 0(), 0(), 0(), s(; x6), x7, x8, x9, x10;) > f(x6, x6, x6, x6, x6, x6, x7, x8, x9, x10;)
f(0(), 0(), 0(), 0(), 0(), 0(), s(; x7), x8, x9, x10;) > f(x7, x7, x7, x7, x7, x7, x7, x8, x9, x10;)
f(0(), 0(), 0(), 0(), 0(), 0(), 0(), s(; x8), x9, x10;) > f(x8, x8, x8, x8, x8, x8, x8, x8, x9, x10;)
f(0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), s(; x9), x10;) > f(x9, x9, x9, x9, x9, x9, x9, x9, x9, x10;)
f(0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), s(; x10);) > f(x10, x10, x10, x10, x10, x10, x10, x10, x10, x10;)
f(0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), 0(), 0();) > 0()
Hurray, we answered YES(?,POLY)