We consider the following Problem:
Strict Trs:
{ a(b(x)) -> b(a(a(x)))
, b(c(x)) -> c(b(b(x)))
, c(a(x)) -> a(c(c(x)))
, u(a(x)) -> x
, v(b(x)) -> x
, w(c(x)) -> x
, a(u(x)) -> x
, b(v(x)) -> x
, c(w(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ a(b(x)) -> b(a(a(x)))
, b(c(x)) -> c(b(b(x)))
, c(a(x)) -> a(c(c(x)))
, u(a(x)) -> x
, v(b(x)) -> x
, w(c(x)) -> x
, a(u(x)) -> x
, b(v(x)) -> x
, c(w(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ w(c(x)) -> x
, c(w(x)) -> x}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a) = {1}, Uargs(b) = {1}, Uargs(c) = {1}, Uargs(u) = {},
Uargs(v) = {}, Uargs(w) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a(x1) = [1 0] x1 + [1]
[1 0] [1]
b(x1) = [1 0] x1 + [0]
[0 0] [1]
c(x1) = [1 0] x1 + [0]
[0 1] [0]
u(x1) = [1 0] x1 + [0]
[0 0] [1]
v(x1) = [1 0] x1 + [1]
[0 0] [1]
w(x1) = [1 0] x1 + [1]
[0 1] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a(b(x)) -> b(a(a(x)))
, b(c(x)) -> c(b(b(x)))
, c(a(x)) -> a(c(c(x)))
, u(a(x)) -> x
, v(b(x)) -> x
, a(u(x)) -> x
, b(v(x)) -> x}
Weak Trs:
{ w(c(x)) -> x
, c(w(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ v(b(x)) -> x
, b(v(x)) -> x}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a) = {1}, Uargs(b) = {1}, Uargs(c) = {1}, Uargs(u) = {},
Uargs(v) = {}, Uargs(w) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a(x1) = [1 0] x1 + [0]
[0 0] [0]
b(x1) = [1 0] x1 + [0]
[0 1] [1]
c(x1) = [1 0] x1 + [0]
[0 1] [0]
u(x1) = [1 0] x1 + [1]
[0 0] [1]
v(x1) = [1 0] x1 + [1]
[0 1] [0]
w(x1) = [1 0] x1 + [1]
[0 1] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a(b(x)) -> b(a(a(x)))
, b(c(x)) -> c(b(b(x)))
, c(a(x)) -> a(c(c(x)))
, u(a(x)) -> x
, a(u(x)) -> x}
Weak Trs:
{ v(b(x)) -> x
, b(v(x)) -> x
, w(c(x)) -> x
, c(w(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ u(a(x)) -> x
, a(u(x)) -> x}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a) = {1}, Uargs(b) = {1}, Uargs(c) = {1}, Uargs(u) = {},
Uargs(v) = {}, Uargs(w) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a(x1) = [1 0] x1 + [0]
[0 1] [1]
b(x1) = [1 0] x1 + [0]
[0 1] [0]
c(x1) = [1 0] x1 + [0]
[0 1] [0]
u(x1) = [1 0] x1 + [1]
[0 1] [0]
v(x1) = [1 0] x1 + [1]
[0 1] [0]
w(x1) = [1 0] x1 + [1]
[0 1] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a(b(x)) -> b(a(a(x)))
, b(c(x)) -> c(b(b(x)))
, c(a(x)) -> a(c(c(x)))}
Weak Trs:
{ u(a(x)) -> x
, a(u(x)) -> x
, v(b(x)) -> x
, b(v(x)) -> x
, w(c(x)) -> x
, c(w(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ a(b(x)) -> b(a(a(x)))
, b(c(x)) -> c(b(b(x)))
, c(a(x)) -> a(c(c(x)))}
Weak Trs:
{ u(a(x)) -> x
, a(u(x)) -> x
, v(b(x)) -> x
, b(v(x)) -> x
, w(c(x)) -> x
, c(w(x)) -> 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:
{ a_0(2) -> 1
, b_0(2) -> 1
, c_0(2) -> 1
, u_0(2) -> 1
, v_0(2) -> 1
, w_0(2) -> 1}
Hurray, we answered YES(?,O(n^1))