We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, p(s(x)) -> x
, eq(0(), 0()) -> true()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, p(s(x)) -> x
, eq(0(), 0()) -> true()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
true() = [0]
[0]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
0() = [0]
[0]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
false() = [3]
[2]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
s(x1) = [0 0] x1 + [0]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, p(s(x)) -> x
, eq(0(), 0()) -> true()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {eq(0(), 0()) -> true()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
true() = [0]
[0]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
0() = [0]
[0]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
false() = [0]
[3]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[0 0] [0 0] [1]
s(x1) = [0 0] x1 + [0]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, p(s(x)) -> x
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{ eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
true() = [0]
[2]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
0() = [3]
[0]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
false() = [0]
[0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [1 0] [2]
s(x1) = [0 0] x1 + [0]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, p(s(x)) -> x
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{ cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [0]
[0 0] [0 0] [0 0] [1]
true() = [0]
[0]
cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [1 1] [0]
0() = [0]
[0]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
false() = [2]
[0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
s(x1) = [0 0] x1 + [2]
[1 0] [3]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, p(s(x)) -> x
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {p(s(x)) -> x}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [0]
[0 0] [0 0] [0 0] [0]
true() = [0]
[0]
cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [1 1] [1]
0() = [3]
[2]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
false() = [3]
[2]
and(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 0] [0 0] [2]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [2]
s(x1) = [0 1] x1 + [1]
[1 0] [3]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
[0 0] [0 0] [0 0] [1]
true() = [0]
[0]
cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
0() = [3]
[0]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
false() = [3]
[0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [1 0] [0]
s(x1) = [0 1] x1 + [0]
[1 0] [2]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, p(0()) -> 0()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {gr(s(x), s(y)) -> gr(x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [0]
[0 0] [0 0] [0 0] [0]
true() = [0]
[1]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [0]
[0 0] [0 0] [0 0] [0]
gr(x1, x2) = [0 0] x1 + [1 1] x2 + [0]
[0 0] [0 0] [1]
0() = [0]
[0]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
false() = [1]
[3]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 0] [3]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
s(x1) = [0 1] x1 + [1]
[1 0] [3]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, p(0()) -> 0()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(s(x), s(y)) -> gr(x, y)
, cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {p(0()) -> 0()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
[0 0] [0 0] [0 0] [0]
true() = [0]
[1]
cond2(x1, x2, x3) = [1 3] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 0] [0 0] [0]
gr(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
0() = [1]
[0]
p(x1) = [1 0] x1 + [1]
[0 1] [0]
false() = [2]
[1]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
s(x1) = [1 0] x1 + [0]
[0 1] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [0]
[0 0] [0 0] [0 0] [1]
true() = [3]
[2]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [0]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
0() = [1]
[1]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
false() = [2]
[1]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [3]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [3]
[1 0] [1 0] [0]
s(x1) = [0 1] x1 + [2]
[1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {gr(0(), x) -> false()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1, 2, 3}, Uargs(gr) = {},
Uargs(p) = {}, Uargs(and) = {1, 2}, Uargs(eq) = {}, Uargs(s) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2, x3) = [1 2] x1 + [1 0] x2 + [1 0] x3 + [0]
[0 0] [0 1] [0 0] [1]
true() = [0]
[1]
cond2(x1, x2, x3) = [1 2] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 1] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [0]
0() = [0]
[0]
p(x1) = [1 0] x1 + [0]
[0 1] [0]
false() = [0]
[0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [0]
eq(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [2]
s(x1) = [1 0] x1 + [1]
[0 1] [3]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We have computed the following dependency pairs
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, eq^#(s(x), 0()) -> c_4()
, eq^#(0(), s(x)) -> c_5()
, p^#(0()) -> c_6()
, gr^#(s(x), s(y)) -> gr^#(x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_9()
, and^#(true(), true()) -> c_10()
, and^#(false(), x) -> c_11()
, and^#(x, false()) -> c_12()
, cond2^#(true(), x, y) -> cond2^#(gr(y, 0()), p(x), p(y))
, eq^#(0(), 0()) -> c_14()
, cond2^#(false(), x, y) ->
cond1^#(and(eq(x, y), gr(x, 0())), x, y)}
We consider the following Problem:
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, eq^#(s(x), s(y)) -> eq^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, eq^#(s(x), 0()) -> c_4()
, eq^#(0(), s(x)) -> c_5()
, p^#(0()) -> c_6()
, gr^#(s(x), s(y)) -> gr^#(x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_9()
, and^#(true(), true()) -> c_10()
, and^#(false(), x) -> c_11()
, and^#(x, false()) -> c_12()
, cond2^#(true(), x, y) -> cond2^#(gr(y, 0()), p(x), p(y))
, eq^#(0(), 0()) -> c_14()
, cond2^#(false(), x, y) ->
cond1^#(and(eq(x, y), gr(x, 0())), x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, cond1(true(), x, y) -> cond2(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond2(true(), x, y) -> cond2(gr(y, 0()), p(x), p(y))
, eq(0(), 0()) -> true()
, cond2(false(), x, y) -> cond1(and(eq(x, y), gr(x, 0())), x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We replace strict/weak-rules by the corresponding usable rules:
Strict Usable Rules:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Usable Rules:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
We consider the following Problem:
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, eq^#(s(x), s(y)) -> eq^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, eq^#(s(x), 0()) -> c_4()
, eq^#(0(), s(x)) -> c_5()
, p^#(0()) -> c_6()
, gr^#(s(x), s(y)) -> gr^#(x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_9()
, and^#(true(), true()) -> c_10()
, and^#(false(), x) -> c_11()
, and^#(x, false()) -> c_12()
, cond2^#(true(), x, y) -> cond2^#(gr(y, 0()), p(x), p(y))
, eq^#(0(), 0()) -> c_14()
, cond2^#(false(), x, y) ->
cond1^#(and(eq(x, y), gr(x, 0())), x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, eq^#(s(x), s(y)) -> eq^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, eq^#(s(x), 0()) -> c_4()
, eq^#(0(), s(x)) -> c_5()
, p^#(0()) -> c_6()
, gr^#(s(x), s(y)) -> gr^#(x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_9()
, and^#(true(), true()) -> c_10()
, and^#(false(), x) -> c_11()
, and^#(x, false()) -> c_12()
, cond2^#(true(), x, y) -> cond2^#(gr(y, 0()), p(x), p(y))
, eq^#(0(), 0()) -> c_14()
, cond2^#(false(), x, y) ->
cond1^#(and(eq(x, y), gr(x, 0())), x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->9:{2} [ YES(?,O(n^1)) ]
|
|->10:{4} [ YES(O(1),O(1)) ]
|
|->11:{5} [ YES(O(1),O(1)) ]
|
`->12:{14} [ YES(O(1),O(1)) ]
->7:{6} [ YES(O(1),O(1)) ]
->6:{7} [ subsumed ]
|
|->13:{1} [ YES(?,O(n^1)) ]
|
`->8:{3} [ YES(O(1),O(1)) ]
->5:{8,15,13} [ YES(O(1),O(1)) ]
->4:{9} [ YES(O(1),O(1)) ]
->3:{10} [ YES(O(1),O(1)) ]
->2:{11} [ YES(O(1),O(1)) ]
->1:{12} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{ 1: gr^#(s(x), 0()) -> c_1()
, 2: eq^#(s(x), s(y)) -> eq^#(x, y)}
WeakDPs DPs:
{ 3: gr^#(0(), x) -> c_3()
, 4: eq^#(s(x), 0()) -> c_4()
, 5: eq^#(0(), s(x)) -> c_5()
, 6: p^#(0()) -> c_6()
, 7: gr^#(s(x), s(y)) -> gr^#(x, y)
, 8: cond1^#(true(), x, y) -> cond2^#(gr(y, 0()), x, y)
, 9: p^#(s(x)) -> c_9()
, 10: and^#(true(), true()) -> c_10()
, 11: and^#(false(), x) -> c_11()
, 12: and^#(x, false()) -> c_12()
, 13: cond2^#(true(), x, y) -> cond2^#(gr(y, 0()), p(x), p(y))
, 14: eq^#(0(), 0()) -> c_14()
, 15: cond2^#(false(), x, y) ->
cond1^#(and(eq(x, y), gr(x, 0())), x, y)}
* Path 9:{2}: YES(?,O(n^1))
-------------------------
We consider the following Problem:
Strict DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
No rule is usable.
We consider the following Problem:
Strict DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ s_0(2) -> 2
, eq^#_0(2, 2) -> 1
, eq^#_1(2, 2) -> 1}
* Path 9:{2}->10:{4}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 9:{2}->11:{5}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 9:{2}->12:{14}: YES(O(1),O(1))
-----------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
Weak DPs: {eq^#(s(x), s(y)) -> eq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 7:{6}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 6:{7}: subsumed
--------------------
This path is subsumed by the proof of paths 6:{7}->13:{1},
6:{7}->8:{3}.
* Path 6:{7}->13:{1}: YES(?,O(n^1))
---------------------------------
We consider the following Problem:
Strict DPs: {gr^#(s(x), 0()) -> c_1()}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {gr^#(s(x), 0()) -> c_1()}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {gr^#(s(x), 0()) -> c_1()}
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
No rule is usable.
We consider the following Problem:
Strict DPs: {gr^#(s(x), 0()) -> c_1()}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ 0_0() -> 2
, s_0(2) -> 2
, gr^#_0(2, 2) -> 1
, c_1_1() -> 1}
* Path 6:{7}->8:{3}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 5:{8,15,13}: YES(O(1),O(1))
--------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 4:{9}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 3:{10}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 2:{11}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 1:{12}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, eq(s(x), s(y)) -> eq(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, eq(s(x), 0()) -> false()
, eq(0(), s(x)) -> false()
, p(0()) -> 0()
, gr(s(x), s(y)) -> gr(x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, eq(0(), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
Hurray, we answered YES(?,O(n^1))