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