We consider the following Problem:
Strict Trs:
{ del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
, f(true(), x, y, z) -> del(.(y, z))
, f(false(), x, y, z) -> .(x, del(.(y, z)))
, =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
, f(true(), x, y, z) -> del(.(y, z))
, f(false(), x, y, z) -> .(x, del(.(y, z)))
, =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {f(false(), x, y, z) -> .(x, del(.(y, z)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(del) = {}, Uargs(.) = {2}, Uargs(f) = {1}, Uargs(=) = {},
Uargs(and) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
del(x1) = [0 0] x1 + [1]
[0 0] [1]
.(x1, x2) = [0 0] x1 + [1 2] x2 + [0]
[0 0] [0 0] [1]
f(x1, x2, x3, x4) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1]
[0 0] [0 1] [0 1] [0 0] [1]
=(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [3]
[0]
nil() = [0]
[0]
u() = [0]
[0]
v() = [0]
[0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
, f(true(), x, y, z) -> del(.(y, z))
, =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))}
Weak Trs: {f(false(), x, y, z) -> .(x, del(.(y, z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {f(true(), x, y, z) -> del(.(y, z))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(del) = {}, Uargs(.) = {2}, Uargs(f) = {1}, Uargs(=) = {},
Uargs(and) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
del(x1) = [0 0] x1 + [1]
[0 0] [1]
.(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
f(x1, x2, x3, x4) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1]
[0 0] [1 1] [1 1] [0 0] [1]
=(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [2]
[0]
false() = [0]
[0]
nil() = [0]
[0]
u() = [0]
[0]
v() = [0]
[0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
, =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))}
Weak Trs:
{ f(true(), x, y, z) -> del(.(y, z))
, f(false(), x, y, z) -> .(x, del(.(y, z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(del) = {}, Uargs(.) = {2}, Uargs(f) = {1}, Uargs(=) = {},
Uargs(and) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
del(x1) = [0 0] x1 + [1]
[0 0] [1]
.(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
f(x1, x2, x3, x4) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [0 0] x4 + [1]
[0 0] [1 0] [1 0] [0 0] [1]
=(x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [0]
[0]
nil() = [0]
[0]
u() = [0]
[0]
v() = [0]
[0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {del(.(x, .(y, z))) -> f(=(x, y), x, y, z)}
Weak Trs:
{ =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))
, f(true(), x, y, z) -> del(.(y, z))
, f(false(), x, y, z) -> .(x, del(.(y, z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {del(.(x, .(y, z))) -> f(=(x, y), x, y, z)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(del) = {}, Uargs(.) = {2}, Uargs(f) = {1}, Uargs(=) = {},
Uargs(and) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
del(x1) = [0 2] x1 + [0]
[0 1] [0]
.(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [1]
f(x1, x2, x3, x4) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [0 2] x4 + [1]
[0 0] [0 0] [0 0] [0 1] [2]
=(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [0]
true() = [1]
[0]
false() = [1]
[0]
nil() = [0]
[0]
u() = [0]
[0]
v() = [0]
[0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Weak Trs:
{ del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
, =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))
, f(true(), x, y, z) -> del(.(y, z))
, f(false(), x, y, z) -> .(x, del(.(y, z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ del(.(x, .(y, z))) -> f(=(x, y), x, y, z)
, =(nil(), nil()) -> true()
, =(.(x, y), nil()) -> false()
, =(nil(), .(y, z)) -> false()
, =(.(x, y), .(u(), v())) -> and(=(x, u()), =(y, v()))
, f(true(), x, y, z) -> del(.(y, z))
, f(false(), x, y, z) -> .(x, del(.(y, z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
Hurray, we answered YES(?,O(n^1))