We consider the following Problem:
Strict Trs:
{ p(f(f(x))) -> q(f(g(x)))
, p(g(g(x))) -> q(g(f(x)))
, q(f(f(x))) -> p(f(g(x)))
, q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ p(f(f(x))) -> q(f(g(x)))
, p(g(g(x))) -> q(g(f(x)))
, q(f(f(x))) -> p(f(g(x)))
, q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {q(g(g(x))) -> p(g(f(x)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(p) = {}, Uargs(f) = {}, Uargs(q) = {}, Uargs(g) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
p(x1) = [0 0] x1 + [1]
[1 0] [1]
f(x1) = [0 0] x1 + [1]
[0 0] [0]
q(x1) = [0 1] x1 + [1]
[0 0] [1]
g(x1) = [0 0] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ p(f(f(x))) -> q(f(g(x)))
, p(g(g(x))) -> q(g(f(x)))
, q(f(f(x))) -> p(f(g(x)))}
Weak Trs: {q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {q(f(f(x))) -> p(f(g(x)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(p) = {}, Uargs(f) = {}, Uargs(q) = {}, Uargs(g) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
p(x1) = [0 0] x1 + [0]
[0 0] [1]
f(x1) = [0 0] x1 + [0]
[0 0] [1]
q(x1) = [0 0] x1 + [1]
[0 1] [0]
g(x1) = [0 0] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ p(f(f(x))) -> q(f(g(x)))
, p(g(g(x))) -> q(g(f(x)))}
Weak Trs:
{ q(f(f(x))) -> p(f(g(x)))
, q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {p(g(g(x))) -> q(g(f(x)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(p) = {}, Uargs(f) = {}, Uargs(q) = {}, Uargs(g) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
p(x1) = [0 1] x1 + [0]
[0 1] [0]
f(x1) = [0 1] x1 + [0]
[0 0] [0]
q(x1) = [0 1] x1 + [0]
[0 1] [1]
g(x1) = [0 1] x1 + [1]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {p(f(f(x))) -> q(f(g(x)))}
Weak Trs:
{ p(g(g(x))) -> q(g(f(x)))
, q(f(f(x))) -> p(f(g(x)))
, q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {p(f(f(x))) -> q(f(g(x)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(p) = {}, Uargs(f) = {}, Uargs(q) = {}, Uargs(g) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
p(x1) = [1 2] x1 + [0]
[0 0] [0]
f(x1) = [1 0] x1 + [3]
[0 0] [1]
q(x1) = [0 2] x1 + [3]
[0 0] [0]
g(x1) = [0 0] x1 + [0]
[0 1] [3]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Weak Trs:
{ p(f(f(x))) -> q(f(g(x)))
, p(g(g(x))) -> q(g(f(x)))
, q(f(f(x))) -> p(f(g(x)))
, q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ p(f(f(x))) -> q(f(g(x)))
, p(g(g(x))) -> q(g(f(x)))
, q(f(f(x))) -> p(f(g(x)))
, q(g(g(x))) -> p(g(f(x)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
Hurray, we answered YES(?,O(n^1))