We consider the following Problem:
Strict Trs:
{ f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x))))
, f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x))))
, f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(f) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
f(x1, x2) = [0 2] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
a() = [0]
[0]
b() = [0]
[2]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x))))}
Weak Trs: {f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x))))}
Weak Trs: {f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 0.
The enriched problem is compatible with the following automaton:
{ f_0(2, 2) -> 1
, a_0() -> 2
, b_0() -> 2}
Hurray, we answered YES(?,O(n^1))