We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(0()) -> 0()
, p(s(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(0()) -> 0()
, p(s(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: {p(0()) -> 0()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 1] [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 1] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[1]
false() = [0]
[0]
cond4(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [0]
[0 0] [3]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [0]
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(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x}
Weak Trs: {p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 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] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[0]
false() = [0]
[2]
cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
and(x1, x2) = [1 0] x1 + [1 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(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x}
Weak Trs:
{ cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 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] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[0]
false() = [0]
[0]
cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
and(x1, x2) = [1 0] x1 + [1 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(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(true(), true()) -> true()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x}
Weak Trs:
{ cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 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] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[0]
false() = [0]
[2]
cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [0]
[0 0] [0]
and(x1, x2) = [1 0] x1 + [1 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(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x}
Weak Trs:
{ and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 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] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[2]
false() = [0]
[0]
cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
p(x1) = [1 0] x1 + [0]
[0 1] [2]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [1]
s(x1) = [1 0] x1 + [2]
[0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{ p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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) -> cond4(gr(y, 0()), x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 + [1]
[0 0] [0 1] [0 1] [1]
true() = [0]
[1]
cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 1] [0 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
cond3(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 1] [0 0] [1]
0() = [0]
[0]
false() = [1]
[2]
cond4(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 1] [0 1] [0 1] [0]
p(x1) = [1 0] x1 + [0]
[0 1] [0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
s(x1) = [1 0] x1 + [0]
[0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{ cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 + [1]
[1 1] [1 1] [1 1] [1]
true() = [2]
[2]
cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [1 1] [1 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [1 0] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [1 1] [0 0] [1]
0() = [0]
[0]
false() = [0]
[1]
cond4(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[1 1] [1 1] [1 1] [0]
p(x1) = [1 0] x1 + [0]
[0 1] [0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [1 0] [0]
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:
{ cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{ cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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() = [2]
[0]
cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [2]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
cond3(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[0]
false() = [0]
[0]
cond4(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
s(x1) = [0 1] x1 + [0]
[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()
, gr(s(x), s(y)) -> gr(x, y)
, and(false(), x) -> false()
, and(x, false()) -> false()}
Weak Trs:
{ cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {and(x, false()) -> false()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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() = [2]
[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 + [1]
[0 0] [0 0] [1]
cond3(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [1]
[0 0] [0 0] [0 0] [1]
0() = [0]
[0]
false() = [0]
[3]
cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [2]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [1]
[1 0] [0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 1] [1]
s(x1) = [0 1] x1 + [0]
[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()
, gr(s(x), s(y)) -> gr(x, y)
, and(false(), x) -> false()}
Weak Trs:
{ and(x, false()) -> false()
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {and(false(), x) -> false()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(cond2) = {1}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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() = [3]
[2]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
cond3(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
[0 0] [0 0] [0 0] [1]
0() = [0]
[0]
false() = [0]
[2]
cond4(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [1 1] x3 + [3]
[0 0] [0 0] [0 0] [1]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
and(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 0] [0 0] [2]
s(x1) = [0 1] x1 + [0]
[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()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ and(false(), x) -> false()
, and(x, false()) -> false()
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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}, Uargs(gr) = {},
Uargs(cond3) = {1, 2}, Uargs(cond4) = {1, 3}, Uargs(p) = {},
Uargs(and) = {1, 2}, 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 0] [0 1] [0]
true() = [3]
[3]
cond2(x1, x2, x3) = [1 2] x1 + [1 0] x2 + [1 0] x3 + [3]
[0 0] [0 0] [0 1] [0]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [0]
cond3(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
[0 0] [0 0] [0 1] [0]
0() = [0]
[0]
false() = [0]
[0]
cond4(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [2]
[0 0] [0 0] [0 1] [0]
p(x1) = [1 0] x1 + [0]
[0 1] [0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [0]
s(x1) = [1 0] x1 + [0]
[0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, and^#(false(), x) -> c_4()
, and^#(x, false()) -> c_5()
, cond3^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
, cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
, cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
, cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_11()
, and^#(true(), true()) -> c_12()
, cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
, cond4^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, p^#(0()) -> c_15()}
We consider the following Problem:
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, and^#(false(), x) -> c_4()
, and^#(x, false()) -> c_5()
, cond3^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
, cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
, cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
, cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_11()
, and^#(true(), true()) -> c_12()
, cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
, cond4^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, p^#(0()) -> c_15()}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, cond3(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1(true(), x, y) -> cond2(gr(x, y), x, y)
, cond2(true(), x, y) -> cond3(gr(x, 0()), x, y)
, cond3(true(), x, y) -> cond3(gr(x, 0()), p(x), y)
, cond2(false(), x, y) -> cond4(gr(y, 0()), x, y)
, p(s(x)) -> x
, and(true(), true()) -> true()
, cond4(true(), x, y) -> cond4(gr(y, 0()), x, p(y))
, cond4(false(), x, y) -> cond1(and(gr(x, 0()), gr(y, 0())), x, y)
, p(0()) -> 0()}
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()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Usable Rules:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
We consider the following Problem:
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, and^#(false(), x) -> c_4()
, and^#(x, false()) -> c_5()
, cond3^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
, cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
, cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
, cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_11()
, and^#(true(), true()) -> c_12()
, cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
, cond4^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, p^#(0()) -> c_15()}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs:
{ gr^#(s(x), 0()) -> c_1()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs:
{ gr^#(0(), x) -> c_3()
, and^#(false(), x) -> c_4()
, and^#(x, false()) -> c_5()
, cond3^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
, cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
, cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
, cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
, p^#(s(x)) -> c_11()
, and^#(true(), true()) -> c_12()
, cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
, cond4^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, p^#(0()) -> c_15()}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->7:{2} [ YES(?,O(n^1)) ]
|
|->9:{1} [ YES(?,O(n^1)) ]
|
`->8:{3} [ YES(O(1),O(1)) ]
->6:{4} [ YES(O(1),O(1)) ]
->5:{5} [ YES(O(1),O(1)) ]
->4:{6,9,8,7,14,13,10} [ YES(O(1),O(1)) ]
->3:{11} [ YES(O(1),O(1)) ]
->2:{12} [ YES(O(1),O(1)) ]
->1:{15} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{ 1: gr^#(s(x), 0()) -> c_1()
, 2: gr^#(s(x), s(y)) -> gr^#(x, y)}
WeakDPs DPs:
{ 3: gr^#(0(), x) -> c_3()
, 4: and^#(false(), x) -> c_4()
, 5: and^#(x, false()) -> c_5()
, 6: cond3^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, 7: cond1^#(true(), x, y) -> cond2^#(gr(x, y), x, y)
, 8: cond2^#(true(), x, y) -> cond3^#(gr(x, 0()), x, y)
, 9: cond3^#(true(), x, y) -> cond3^#(gr(x, 0()), p(x), y)
, 10: cond2^#(false(), x, y) -> cond4^#(gr(y, 0()), x, y)
, 11: p^#(s(x)) -> c_11()
, 12: and^#(true(), true()) -> c_12()
, 13: cond4^#(true(), x, y) -> cond4^#(gr(y, 0()), x, p(y))
, 14: cond4^#(false(), x, y) ->
cond1^#(and(gr(x, 0()), gr(y, 0())), x, y)
, 15: p^#(0()) -> c_15()}
* Path 7:{2}: YES(?,O(n^1))
-------------------------
We consider the following Problem:
Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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), 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:
{ s_0(2) -> 2
, gr^#_0(2, 2) -> 1
, gr^#_1(2, 2) -> 1}
* Path 7:{2}->9:{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()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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 7:{2}->8:{3}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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 6:{4}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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 5:{5}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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:{6,9,8,7,14,13,10}: YES(O(1),O(1))
-----------------------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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:{11}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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:{12}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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:{15}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ gr(0(), x) -> false()
, and(false(), x) -> false()
, and(x, false()) -> false()
, p(s(x)) -> x
, and(true(), true()) -> true()
, p(0()) -> 0()}
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))