We consider the following Problem:
Strict Trs:
{ cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, cond2(true(), x, y) -> cond1(y, y)
, cond2(false(), x, y) -> cond1(p(x), y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()
, neq(s(x), s(y)) -> neq(x, y)
, p(0()) -> 0()
, p(s(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We consider the following Problem:
Strict Trs:
{ cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, cond2(true(), x, y) -> cond1(y, y)
, cond2(false(), x, y) -> cond1(p(x), y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()
, neq(s(x), s(y)) -> neq(x, y)
, p(0()) -> 0()
, p(s(x)) -> x}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 1] [1]
s(x1) = [1 0] x1 + [0]
[0 0] [0]
cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [1]
[0 0] [0 0] [0 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [0]
[0]
p(x1) = [1 0] x1 + [0]
[0 0] [1]
0() = [0]
[0]
neq(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:
{ cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, cond2(true(), x, y) -> cond1(y, y)
, cond2(false(), x, y) -> cond1(p(x), y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, neq(s(x), s(y)) -> neq(x, y)
, p(0()) -> 0()
, p(s(x)) -> x}
Weak Trs:
{ neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {cond2(false(), x, y) -> cond1(p(x), y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 1] [1]
s(x1) = [1 0] x1 + [0]
[0 0] [0]
cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [1]
[0 0] [0 0] [0 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [1]
[0]
p(x1) = [1 0] x1 + [0]
[0 0] [1]
0() = [1]
[0]
neq(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:
{ cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, neq(s(x), s(y)) -> neq(x, y)
, p(0()) -> 0()
, p(s(x)) -> x}
Weak Trs:
{ cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
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(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 1] [1]
s(x1) = [1 0] x1 + [0]
[0 0] [0]
cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [1]
[0 0] [0 0] [0 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [1]
[0]
p(x1) = [1 0] x1 + [1]
[0 0] [1]
0() = [0]
[0]
neq(x1, x2) = [1 0] x1 + [1 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:
{ cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, neq(s(x), s(y)) -> neq(x, y)
, p(s(x)) -> x}
Weak Trs:
{ p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 1] [1]
s(x1) = [1 0] x1 + [0]
[0 0] [0]
cond2(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [2]
[0]
p(x1) = [1 0] x1 + [1]
[0 0] [1]
0() = [2]
[0]
neq(x1, x2) = [1 0] x1 + [1 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:
{ cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), 0()) -> true()
, gr(s(x), s(y)) -> gr(x, y)
, neq(s(x), s(y)) -> neq(x, y)
, p(s(x)) -> x}
Weak Trs:
{ cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {gr(s(x), 0()) -> true()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 1] x1 + [0 0] x2 + [3]
[0 0] [1 1] [1]
s(x1) = [0 0] x1 + [0]
[1 0] [0]
cond2(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [0 0] x3 + [1]
[0 0] [0 0] [1 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [1]
[3]
p(x1) = [1 1] x1 + [1]
[0 0] [1]
0() = [3]
[0]
neq(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [1 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)
, neq(s(x), s(y)) -> neq(x, y)
, p(s(x)) -> x}
Weak Trs:
{ gr(s(x), 0()) -> true()
, cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
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(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 1] x1 + [0 0] x2 + [2]
[0 0] [1 1] [1]
s(x1) = [0 1] x1 + [0]
[1 0] [2]
cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [0 0] x3 + [1]
[0 0] [0 0] [1 1] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [3]
[0]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
0() = [0]
[0]
neq(x1, x2) = [0 0] x1 + [0 0] x2 + [3]
[0 0] [1 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)
, neq(s(x), s(y)) -> neq(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {neq(s(x), s(y)) -> neq(x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(cond1) = {1}, Uargs(s) = {}, Uargs(cond2) = {1},
Uargs(gr) = {}, Uargs(p) = {}, Uargs(neq) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
cond1(x1, x2) = [1 1] x1 + [0 0] x2 + [2]
[0 0] [0 0] [1]
s(x1) = [0 1] x1 + [2]
[1 0] [2]
cond2(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [0 0] x3 + [2]
[0 0] [0 0] [0 0] [1]
gr(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
true() = [0]
[0]
false() = [2]
[0]
p(x1) = [0 1] x1 + [0]
[1 0] [0]
0() = [1]
[1]
neq(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ neq(s(x), s(y)) -> neq(x, y)
, p(s(x)) -> x
, gr(s(x), 0()) -> true()
, cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We consider the following Problem:
Strict Trs:
{ cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ neq(s(x), s(y)) -> neq(x, y)
, p(s(x)) -> x
, gr(s(x), 0()) -> true()
, cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We have computed the following dependency pairs
Strict DPs:
{ cond2^#(true(), x, y) -> cond1^#(y, y)
, gr^#(0(), x) -> c_2()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak DPs:
{ neq^#(s(x), s(y)) -> neq^#(x, y)
, p^#(s(x)) -> c_5()
, gr^#(s(x), 0()) -> c_6()
, cond1^#(s(x), y) -> cond2^#(gr(s(x), y), s(x), y)
, p^#(0()) -> c_8()
, cond2^#(false(), x, y) -> cond1^#(p(x), y)
, neq^#(0(), 0()) -> c_10()
, neq^#(0(), s(x)) -> c_11()
, neq^#(s(x), 0()) -> c_12()}
We consider the following Problem:
Strict DPs:
{ cond2^#(true(), x, y) -> cond1^#(y, y)
, gr^#(0(), x) -> c_2()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ cond2(true(), x, y) -> cond1(y, y)
, gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs:
{ neq^#(s(x), s(y)) -> neq^#(x, y)
, p^#(s(x)) -> c_5()
, gr^#(s(x), 0()) -> c_6()
, cond1^#(s(x), y) -> cond2^#(gr(s(x), y), s(x), y)
, p^#(0()) -> c_8()
, cond2^#(false(), x, y) -> cond1^#(p(x), y)
, neq^#(0(), 0()) -> c_10()
, neq^#(0(), s(x)) -> c_11()
, neq^#(s(x), 0()) -> c_12()}
Weak Trs:
{ neq(s(x), s(y)) -> neq(x, y)
, p(s(x)) -> x
, gr(s(x), 0()) -> true()
, cond1(s(x), y) -> cond2(gr(s(x), y), s(x), y)
, p(0()) -> 0()
, cond2(false(), x, y) -> cond1(p(x), y)
, neq(0(), 0()) -> false()
, neq(0(), s(x)) -> true()
, neq(s(x), 0()) -> true()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We replace strict/weak-rules by the corresponding usable rules:
Strict Usable Rules:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Usable Rules:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
We consider the following Problem:
Strict DPs:
{ cond2^#(true(), x, y) -> cond1^#(y, y)
, gr^#(0(), x) -> c_2()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs:
{ neq^#(s(x), s(y)) -> neq^#(x, y)
, p^#(s(x)) -> c_5()
, gr^#(s(x), 0()) -> c_6()
, cond1^#(s(x), y) -> cond2^#(gr(s(x), y), s(x), y)
, p^#(0()) -> c_8()
, cond2^#(false(), x, y) -> cond1^#(p(x), y)
, neq^#(0(), 0()) -> c_10()
, neq^#(0(), s(x)) -> c_11()
, neq^#(s(x), 0()) -> c_12()}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We consider the following Problem:
Strict DPs:
{ cond2^#(true(), x, y) -> cond1^#(y, y)
, gr^#(0(), x) -> c_2()
, gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs:
{ neq^#(s(x), s(y)) -> neq^#(x, y)
, p^#(s(x)) -> c_5()
, gr^#(s(x), 0()) -> c_6()
, cond1^#(s(x), y) -> cond2^#(gr(s(x), y), s(x), y)
, p^#(0()) -> c_8()
, cond2^#(false(), x, y) -> cond1^#(p(x), y)
, neq^#(0(), 0()) -> c_10()
, neq^#(0(), s(x)) -> c_11()
, neq^#(s(x), 0()) -> c_12()}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We use following congruence DG for path analysis
->10:{1,7,9} [ YES(?,O(n^2)) ]
->7:{3} [ YES(?,O(n^1)) ]
|
|->9:{2} [ YES(?,O(n^1)) ]
|
`->8:{6} [ YES(O(1),O(1)) ]
->3:{4} [ subsumed ]
|
|->4:{10} [ YES(O(1),O(1)) ]
|
|->5:{11} [ YES(O(1),O(1)) ]
|
`->6:{12} [ YES(O(1),O(1)) ]
->2:{5} [ YES(O(1),O(1)) ]
->1:{8} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{ 1: cond2^#(true(), x, y) -> cond1^#(y, y)
, 2: gr^#(0(), x) -> c_2()
, 3: gr^#(s(x), s(y)) -> gr^#(x, y)}
WeakDPs DPs:
{ 4: neq^#(s(x), s(y)) -> neq^#(x, y)
, 5: p^#(s(x)) -> c_5()
, 6: gr^#(s(x), 0()) -> c_6()
, 7: cond1^#(s(x), y) -> cond2^#(gr(s(x), y), s(x), y)
, 8: p^#(0()) -> c_8()
, 9: cond2^#(false(), x, y) -> cond1^#(p(x), y)
, 10: neq^#(0(), 0()) -> c_10()
, 11: neq^#(0(), s(x)) -> c_11()
, 12: neq^#(s(x), 0()) -> c_12()}
* Path 10:{1,7,9}: YES(?,O(n^2))
------------------------------
We consider the following Problem:
Strict DPs: {cond2^#(true(), x, y) -> cond1^#(y, y)}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We consider the following Problem:
Strict DPs: {cond2^#(true(), x, y) -> cond1^#(y, y)}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
We consider the following Problem:
Strict DPs: {cond2^#(true(), x, y) -> cond1^#(y, y)}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
No rule is usable.
We consider the following Problem:
Strict DPs: {cond2^#(true(), x, y) -> cond1^#(y, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Proof:
The following argument positions are usable:
Uargs(cond2^#) = {}, Uargs(cond1^#) = {}
We have the following restricted polynomial interpretation:
Interpretation Functions:
[true]() = 2
[cond2^#](x1, x2, x3) = x1^2 + 2*x3^2
[cond1^#](x1, x2) = x1^2
* Path 7:{3}: YES(?,O(n^1))
-------------------------
We consider the following Problem:
Strict DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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:{3}->9:{2}: YES(?,O(n^1))
--------------------------------
We consider the following Problem:
Strict DPs: {gr^#(0(), x) -> c_2()}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {gr^#(0(), x) -> c_2()}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {gr^#(0(), x) -> c_2()}
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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^#(0(), x) -> c_2()}
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:
{ s_0(2) -> 2
, 0_0() -> 2
, gr^#_0(2, 2) -> 1
, c_2_1() -> 1}
* Path 7:{3}->8:{6}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {gr^#(s(x), s(y)) -> gr^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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 3:{4}: subsumed
--------------------
This path is subsumed by the proof of paths 3:{4}->6:{12},
3:{4}->5:{11},
3:{4}->4:{10}.
* Path 3:{4}->4:{10}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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: {neq^#(s(x), s(y)) -> neq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 3:{4}->5:{11}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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: {neq^#(s(x), s(y)) -> neq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 3:{4}->6:{12}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak DPs: {neq^#(s(x), s(y)) -> neq^#(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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: {neq^#(s(x), s(y)) -> neq^#(x, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 2:{5}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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:{8}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> true()
, p(0()) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Strict Trs:
{ gr(0(), x) -> false()
, gr(s(x), s(y)) -> gr(x, y)}
Weak Trs:
{ p(s(x)) -> x
, gr(s(x), 0()) -> 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^2))