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