We consider the following Problem:
Strict Trs:
{ from(X) -> cons(X, n__from(n__s(X)))
, head(cons(X, XS)) -> X
, 2nd(cons(X, XS)) -> head(activate(XS))
, take(0(), XS) -> nil()
, take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
, sel(0(), cons(X, XS)) -> X
, sel(s(N), cons(X, XS)) -> sel(N, activate(XS))
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, take(X1, X2) -> n__take(X1, X2)
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
Arguments of following rules are not normal-forms:
{ take(s(N), cons(X, XS)) -> cons(X, n__take(N, activate(XS)))
, sel(s(N), cons(X, XS)) -> sel(N, activate(XS))}
All above mentioned rules can be savely removed.
We consider the following Problem:
Strict Trs:
{ from(X) -> cons(X, n__from(n__s(X)))
, head(cons(X, XS)) -> X
, 2nd(cons(X, XS)) -> head(activate(XS))
, take(0(), XS) -> nil()
, sel(0(), cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, take(X1, X2) -> n__take(X1, X2)
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(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:
{ take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {1}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 1] x1 + [1]
[0 0] [1]
cons(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
n__from(x1) = [0 0] x1 + [0]
[1 1] [0]
n__s(x1) = [0 0] x1 + [0]
[1 1] [0]
head(x1) = [1 0] x1 + [0]
[0 0] [1]
2nd(x1) = [1 0] x1 + [0]
[0 0] [1]
activate(x1) = [1 1] x1 + [0]
[0 0] [0]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 1] [0 1] [1]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [1]
[0 0] [1]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
sel(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ from(X) -> cons(X, n__from(n__s(X)))
, head(cons(X, XS)) -> X
, 2nd(cons(X, XS)) -> head(activate(XS))
, sel(0(), cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X}
Weak Trs:
{ take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {from(X) -> cons(X, n__from(n__s(X)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {1}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 1] x1 + [2]
[0 0] [2]
cons(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
n__from(x1) = [0 0] x1 + [0]
[1 1] [0]
n__s(x1) = [0 0] x1 + [0]
[1 1] [0]
head(x1) = [1 0] x1 + [0]
[0 0] [1]
2nd(x1) = [1 0] x1 + [0]
[0 0] [1]
activate(x1) = [1 1] x1 + [0]
[0 0] [0]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [1]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [0]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
sel(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ head(cons(X, XS)) -> X
, 2nd(cons(X, XS)) -> head(activate(XS))
, sel(0(), cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X}
Weak Trs:
{ from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ from(X) -> n__from(X)
, s(X) -> n__s(X)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {1}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 0] x1 + [1]
[0 0] [1]
cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
n__from(x1) = [1 0] x1 + [0]
[0 0] [0]
n__s(x1) = [1 0] x1 + [0]
[0 0] [0]
head(x1) = [1 0] x1 + [1]
[0 0] [1]
2nd(x1) = [1 0] x1 + [1]
[0 0] [1]
activate(x1) = [1 0] x1 + [0]
[0 0] [1]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [1]
[0 0] [1]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
sel(x1, x2) = [0 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:
{ head(cons(X, XS)) -> X
, 2nd(cons(X, XS)) -> head(activate(XS))
, sel(0(), cons(X, XS)) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X}
Weak Trs:
{ from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {head(cons(X, XS)) -> X}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {1}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 0] x1 + [0]
[0 1] [3]
cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
n__from(x1) = [1 0] x1 + [0]
[0 0] [0]
n__s(x1) = [1 0] x1 + [0]
[0 0] [0]
head(x1) = [1 0] x1 + [1]
[0 1] [1]
2nd(x1) = [1 0] x1 + [1]
[0 1] [1]
activate(x1) = [1 0] x1 + [0]
[0 0] [1]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
sel(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ 2nd(cons(X, XS)) -> head(activate(XS))
, sel(0(), cons(X, XS)) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X}
Weak Trs:
{ head(cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {sel(0(), cons(X, XS)) -> X}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {1}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 0] x1 + [0]
[0 1] [0]
cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [0]
n__from(x1) = [1 0] x1 + [0]
[0 0] [0]
n__s(x1) = [1 0] x1 + [1]
[1 0] [0]
head(x1) = [1 0] x1 + [1]
[0 1] [0]
2nd(x1) = [1 0] x1 + [1]
[0 0] [0]
activate(x1) = [1 0] x1 + [0]
[0 0] [1]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[1 0] [1 0] [1]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [1]
[1 0] [1]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
sel(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
[0 0] [0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ 2nd(cons(X, XS)) -> head(activate(XS))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X}
Weak Trs:
{ sel(0(), cons(X, XS)) -> X
, head(cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {activate(X) -> X}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {1}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 0] x1 + [0]
[0 1] [0]
cons(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [0]
n__from(x1) = [1 0] x1 + [0]
[0 0] [0]
n__s(x1) = [1 0] x1 + [0]
[0 0] [1]
head(x1) = [1 0] x1 + [1]
[0 1] [1]
2nd(x1) = [1 0] x1 + [0]
[0 0] [1]
activate(x1) = [1 0] x1 + [1]
[0 1] [3]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 1] [2]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
sel(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
[0 0] [0 1] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ 2nd(cons(X, XS)) -> head(activate(XS))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))}
Weak Trs:
{ activate(X) -> X
, sel(0(), cons(X, XS)) -> X
, head(cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ 2nd(cons(X, XS)) -> head(activate(XS))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))}
Weak Trs:
{ activate(X) -> X
, sel(0(), cons(X, XS)) -> X
, head(cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We have computed the following dependency pairs
Strict DPs:
{ 2nd^#(cons(X, XS)) -> head^#(activate(XS))
, activate^#(n__from(X)) -> from^#(activate(X))
, activate^#(n__s(X)) -> s^#(activate(X))
, activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
Weak DPs:
{ activate^#(X) -> c_5()
, sel^#(0(), cons(X, XS)) -> c_6()
, head^#(cons(X, XS)) -> c_7()
, from^#(X) -> c_8()
, s^#(X) -> c_9()
, from^#(X) -> c_10()
, take^#(0(), XS) -> c_11()
, take^#(X1, X2) -> c_12()}
We consider the following Problem:
Strict DPs:
{ 2nd^#(cons(X, XS)) -> head^#(activate(XS))
, activate^#(n__from(X)) -> from^#(activate(X))
, activate^#(n__s(X)) -> s^#(activate(X))
, activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
Strict Trs:
{ 2nd(cons(X, XS)) -> head(activate(XS))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))}
Weak DPs:
{ activate^#(X) -> c_5()
, sel^#(0(), cons(X, XS)) -> c_6()
, head^#(cons(X, XS)) -> c_7()
, from^#(X) -> c_8()
, s^#(X) -> c_9()
, from^#(X) -> c_10()
, take^#(0(), XS) -> c_11()
, take^#(X1, X2) -> c_12()}
Weak Trs:
{ activate(X) -> X
, sel(0(), cons(X, XS)) -> X
, head(cons(X, XS)) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We replace strict/weak-rules by the corresponding usable rules:
Strict Usable Rules:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))}
Weak Usable Rules:
{ activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
We consider the following Problem:
Strict DPs:
{ 2nd^#(cons(X, XS)) -> head^#(activate(XS))
, activate^#(n__from(X)) -> from^#(activate(X))
, activate^#(n__s(X)) -> s^#(activate(X))
, activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
Strict Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))}
Weak DPs:
{ activate^#(X) -> c_5()
, sel^#(0(), cons(X, XS)) -> c_6()
, head^#(cons(X, XS)) -> c_7()
, from^#(X) -> c_8()
, s^#(X) -> c_9()
, from^#(X) -> c_10()
, take^#(0(), XS) -> c_11()
, take^#(X1, X2) -> c_12()}
Weak Trs:
{ activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
Dependency Pairs:
{ activate^#(n__from(X)) -> from^#(activate(X))
, activate^#(n__s(X)) -> s^#(activate(X))
, activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
TRS Component:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))}
Interpretation of constant growth:
----------------------------------
The following argument positions are usable:
Uargs(from) = {1}, Uargs(cons) = {}, Uargs(n__from) = {},
Uargs(n__s) = {}, Uargs(head) = {}, Uargs(2nd) = {},
Uargs(activate) = {}, Uargs(take) = {1, 2}, Uargs(s) = {1},
Uargs(n__take) = {}, Uargs(sel) = {}, Uargs(2nd^#) = {},
Uargs(head^#) = {1}, Uargs(activate^#) = {}, Uargs(from^#) = {1},
Uargs(s^#) = {1}, Uargs(take^#) = {1, 2}, Uargs(sel^#) = {}
We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
from(x1) = [1 0] x1 + [2]
[0 1] [2]
cons(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 1] [0 0] [2]
n__from(x1) = [1 0] x1 + [2]
[0 1] [2]
n__s(x1) = [1 0] x1 + [0]
[0 1] [2]
head(x1) = [0 0] x1 + [0]
[0 0] [0]
2nd(x1) = [0 0] x1 + [0]
[0 0] [0]
activate(x1) = [2 2] x1 + [0]
[0 2] [1]
take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [3]
0() = [0]
[0]
nil() = [0]
[0]
s(x1) = [1 0] x1 + [1]
[0 1] [2]
n__take(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
sel(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
2nd^#(x1) = [2 2] x1 + [0]
[0 0] [0]
head^#(x1) = [1 0] x1 + [0]
[2 1] [0]
activate^#(x1) = [2 2] x1 + [0]
[0 0] [0]
from^#(x1) = [1 0] x1 + [0]
[0 0] [0]
s^#(x1) = [1 0] x1 + [0]
[0 0] [0]
take^#(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [0]
c_5() = [0]
[0]
sel^#(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
c_6() = [0]
[0]
c_7() = [0]
[0]
c_8() = [0]
[0]
c_9() = [0]
[0]
c_10() = [0]
[0]
c_11() = [0]
[0]
c_12() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict DPs: {2nd^#(cons(X, XS)) -> head^#(activate(XS))}
Weak DPs:
{ activate^#(n__from(X)) -> from^#(activate(X))
, activate^#(n__s(X)) -> s^#(activate(X))
, activate^#(n__take(X1, X2)) -> take^#(activate(X1), activate(X2))
, activate^#(X) -> c_5()
, sel^#(0(), cons(X, XS)) -> c_6()
, head^#(cons(X, XS)) -> c_7()
, from^#(X) -> c_8()
, s^#(X) -> c_9()
, from^#(X) -> c_10()
, take^#(0(), XS) -> c_11()
, take^#(X1, X2) -> c_12()}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->11:{1} [ YES(O(1),O(1)) ]
|
`->12:{7} [ YES(O(1),O(1)) ]
->8:{2} [ subsumed ]
|
|->9:{8} [ YES(O(1),O(1)) ]
|
`->10:{10} [ YES(O(1),O(1)) ]
->6:{3} [ subsumed ]
|
`->7:{9} [ YES(O(1),O(1)) ]
->3:{4} [ subsumed ]
|
|->4:{11} [ YES(O(1),O(1)) ]
|
`->5:{12} [ YES(O(1),O(1)) ]
->2:{5} [ YES(O(1),O(1)) ]
->1:{6} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))}
WeakDPs DPs:
{ 2: activate^#(n__from(X)) -> from^#(activate(X))
, 3: activate^#(n__s(X)) -> s^#(activate(X))
, 4: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))
, 5: activate^#(X) -> c_5()
, 6: sel^#(0(), cons(X, XS)) -> c_6()
, 7: head^#(cons(X, XS)) -> c_7()
, 8: from^#(X) -> c_8()
, 9: s^#(X) -> c_9()
, 10: from^#(X) -> c_10()
, 11: take^#(0(), XS) -> c_11()
, 12: take^#(X1, X2) -> c_12()}
* Path 11:{1}: YES(O(1),O(1))
---------------------------
We consider the following Problem:
Strict DPs: {2nd^#(cons(X, XS)) -> head^#(activate(XS))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))
together with the congruence-graph
->1:{1} Noncyclic, trivial, SCC
Here dependency-pairs are as follows:
Strict DPs:
{1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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 11:{1}->12:{7}: YES(O(1),O(1))
-----------------------------------
We consider the following Problem:
Weak DPs: {2nd^#(cons(X, XS)) -> head^#(activate(XS))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: 2nd^#(cons(X, XS)) -> head^#(activate(XS))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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 8:{2}: subsumed
--------------------
This path is subsumed by the proof of paths 8:{2}->10:{10},
8:{2}->9:{8}.
* Path 8:{2}->9:{8}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {activate^#(n__from(X)) -> from^#(activate(X))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: activate^#(n__from(X)) -> from^#(activate(X))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: activate^#(n__from(X)) -> from^#(activate(X))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: activate^#(n__from(X)) -> from^#(activate(X))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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 8:{2}->10:{10}: YES(O(1),O(1))
-----------------------------------
We consider the following Problem:
Weak DPs: {activate^#(n__from(X)) -> from^#(activate(X))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: activate^#(n__from(X)) -> from^#(activate(X))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: activate^#(n__from(X)) -> from^#(activate(X))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: activate^#(n__from(X)) -> from^#(activate(X))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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:{3}: subsumed
--------------------
This path is subsumed by the proof of paths 6:{3}->7:{9}.
* Path 6:{3}->7:{9}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {activate^#(n__s(X)) -> s^#(activate(X))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: activate^#(n__s(X)) -> s^#(activate(X))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: activate^#(n__s(X)) -> s^#(activate(X))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: activate^#(n__s(X)) -> s^#(activate(X))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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:{4}: subsumed
--------------------
This path is subsumed by the proof of paths 3:{4}->5:{12},
3:{4}->4:{11}.
* Path 3:{4}->4:{11}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Weak DPs:
{activate^#(n__take(X1, X2)) -> take^#(activate(X1), activate(X2))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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:{4}->5:{12}: YES(O(1),O(1))
----------------------------------
We consider the following Problem:
Weak DPs:
{activate^#(n__take(X1, X2)) -> take^#(activate(X1), activate(X2))}
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: activate^#(n__take(X1, X2)) ->
take^#(activate(X1), activate(X2))}
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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:{5}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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:{6}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, activate(n__take(X1, X2)) -> take(activate(X1), activate(X2))
, activate(X) -> X
, from(X) -> n__from(X)
, s(X) -> n__s(X)
, from(X) -> cons(X, n__from(n__s(X)))
, take(0(), XS) -> nil()
, take(X1, X2) -> n__take(X1, X2)}
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))