We consider the following Problem:
Strict Trs:
{ c(z, x, a()) -> f(b(b(f(z), z), x))
, b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))
, f(c(c(z, a(), a()), x, a())) -> z}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
Arguments of following rules are not normal-forms:
{f(c(c(z, a(), a()), x, a())) -> z}
All above mentioned rules can be savely removed.
We consider the following Problem:
Strict Trs:
{ c(z, x, a()) -> f(b(b(f(z), z), x))
, b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {c(z, x, a()) -> f(b(b(f(z), z), x))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(c) = {}, Uargs(f) = {1}, Uargs(b) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
c(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1]
[0 0] [0 0] [0 0] [1]
a() = [0]
[0]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
b(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [1 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))}
Weak Trs: {c(z, x, a()) -> f(b(b(f(z), z), x))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs: {b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))}
Weak Trs: {c(z, x, a()) -> f(b(b(f(z), z), x))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The following argument positions are usable:
Uargs(c) = {}, Uargs(f) = {1}, Uargs(b) = {1}
We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
c(x1, x2, x3) = [2 0] x1 + [1 0] x2 + [0 2] x3 + [0]
[0 0] [0 1] [0 0] [0]
a() = [0]
[2]
f(x1) = [1 0] x1 + [0]
[0 0] [0]
b(x1, x2) = [1 2] x1 + [1 0] x2 + [1]
[0 0] [0 0] [0]
Hurray, we answered YES(?,O(n^1))