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)