We consider the following Problem:
Strict Trs:
{ :(x, x) -> e()
, :(x, e()) -> x
, i(:(x, y)) -> :(y, x)
, :(:(x, y), z) -> :(x, :(z, i(y)))
, :(e(), x) -> i(x)
, i(i(x)) -> x
, i(e()) -> e()
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ :(x, x) -> e()
, :(x, e()) -> x
, i(:(x, y)) -> :(y, x)
, :(:(x, y), z) -> :(x, :(z, i(y)))
, :(e(), x) -> i(x)
, i(i(x)) -> x
, i(e()) -> e()
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ :(x, x) -> e()
, :(x, e()) -> x}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(:) = {1, 2}, Uargs(i) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
:(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 1] [0 0] [1]
e() = [0]
[0]
i(x1) = [1 0] x1 + [0]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ i(:(x, y)) -> :(y, x)
, :(:(x, y), z) -> :(x, :(z, i(y)))
, :(e(), x) -> i(x)
, i(i(x)) -> x
, i(e()) -> e()
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)}
Weak Trs:
{ :(x, x) -> e()
, :(x, e()) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {i(e()) -> e()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(:) = {1, 2}, Uargs(i) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
:(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 0] [0]
e() = [0]
[0]
i(x1) = [1 0] x1 + [1]
[1 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ i(:(x, y)) -> :(y, x)
, :(:(x, y), z) -> :(x, :(z, i(y)))
, :(e(), x) -> i(x)
, i(i(x)) -> x
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)}
Weak Trs:
{ i(e()) -> e()
, :(x, x) -> e()
, :(x, e()) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ :(e(), x) -> i(x)
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(:) = {1, 2}, Uargs(i) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
:(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [1 0] [0]
e() = [2]
[0]
i(x1) = [1 0] x1 + [0]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ i(:(x, y)) -> :(y, x)
, :(:(x, y), z) -> :(x, :(z, i(y)))
, i(i(x)) -> x}
Weak Trs:
{ :(e(), x) -> i(x)
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)
, i(e()) -> e()
, :(x, x) -> e()
, :(x, e()) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {i(i(x)) -> x}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(:) = {1, 2}, Uargs(i) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
:(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 1] [1 1] [3]
e() = [0]
[0]
i(x1) = [1 0] x1 + [1]
[0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ i(:(x, y)) -> :(y, x)
, :(:(x, y), z) -> :(x, :(z, i(y)))}
Weak Trs:
{ i(i(x)) -> x
, :(e(), x) -> i(x)
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)
, i(e()) -> e()
, :(x, x) -> e()
, :(x, e()) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {i(:(x, y)) -> :(y, x)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(:) = {1, 2}, Uargs(i) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
:(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [0]
e() = [2]
[0]
i(x1) = [1 0] x1 + [1]
[0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {:(:(x, y), z) -> :(x, :(z, i(y)))}
Weak Trs:
{ i(:(x, y)) -> :(y, x)
, i(i(x)) -> x
, :(e(), x) -> i(x)
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)
, i(e()) -> e()
, :(x, x) -> e()
, :(x, e()) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs: {:(:(x, y), z) -> :(x, :(z, i(y)))}
Weak Trs:
{ i(:(x, y)) -> :(y, x)
, i(i(x)) -> x
, :(e(), x) -> i(x)
, :(x, :(y, i(x))) -> i(y)
, :(x, :(y, :(i(x), z))) -> :(i(z), y)
, :(i(x), :(y, x)) -> i(y)
, :(i(x), :(y, :(x, z))) -> :(i(z), y)
, i(e()) -> e()
, :(x, x) -> e()
, :(x, e()) -> 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:
{ :_0(2, 2) -> 1
, e_0() -> 1
, e_0() -> 2
, i_0(2) -> 1}
Hurray, we answered YES(?,O(n^1))