We consider the following Problem:
Strict Trs:
{ a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(X1, X2)
, a__from(X) -> from(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(X1, X2)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
mark(x1) = [0 0] x1 + [1]
[0 0] [1]
false() = [0]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1]
[0 0] [0 0] [0 0] [1]
a__add(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [1]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [1]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__from(x1) = [0 0] x1 + [0]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__from(X) -> from(X)}
Weak Trs:
{ a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {a__from(X) -> from(X)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
mark(x1) = [0 0] x1 + [1]
[0 0] [1]
false() = [0]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1]
[1 0] [0 0] [0 0] [1]
a__add(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[1 0] [0 0] [0]
0() = [1]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [0]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [1]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__from(x1) = [0 0] x1 + [1]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(X1, X2)}
Weak Trs:
{ a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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:
{ a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 3] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
mark(x1) = [0 0] x1 + [0]
[0 0] [1]
false() = [0]
[0]
a__if(x1, x2, x3) = [1 3] x1 + [0 0] x2 + [0 0] x3 + [1]
[0 0] [0 0] [0 0] [1]
a__add(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [1]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [1]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__from(x1) = [0 0] x1 + [0]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(X1, X2)}
Weak Trs:
{ a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {mark(cons(X1, X2)) -> cons(X1, X2)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 0] x2 + [3]
[0 0] [0 1] [1]
true() = [2]
[0]
mark(x1) = [0 0] x1 + [2]
[0 1] [1]
false() = [2]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1]
[1 0] [0 1] [0 1] [1]
a__add(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 1] [1]
0() = [2]
[1]
s(x1) = [0 0] x1 + [0]
[0 0] [1]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [1]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__from(x1) = [0 0] x1 + [0]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[1 0] [0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))}
Weak Trs:
{ mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {a__from(X) -> cons(X, from(s(X)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 1] [1]
true() = [1]
[0]
mark(x1) = [0 0] x1 + [1]
[0 1] [1]
false() = [1]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 1] [0 1] [1]
a__add(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 1] [1]
0() = [1]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [1]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [1]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__from(x1) = [0 0] x1 + [1]
[0 1] [1]
from(x1) = [0 0] x1 + [0]
[0 1] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))}
Weak Trs:
{ a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
true() = [1]
[1]
mark(x1) = [0 0] x1 + [1]
[0 0] [1]
false() = [0]
[1]
a__if(x1, x2, x3) = [1 2] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
a__add(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
0() = [1]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [1]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
a__from(x1) = [0 0] x1 + [0]
[0 0] [1]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))}
Weak Trs:
{ a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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:
{mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 1] x2 + [1]
[0 1] [0 1] [1]
true() = [0]
[0]
mark(x1) = [0 1] x1 + [0]
[0 1] [0]
false() = [0]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 1] x2 + [0 1] x3 + [1]
[0 1] [0 1] [0 1] [1]
a__add(x1, x2) = [1 0] x1 + [0 1] x2 + [1]
[0 1] [0 1] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[1 0] [0]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 1] [0 1] [2]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [1 0] [0]
first(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
a__from(x1) = [1 0] x1 + [0]
[1 0] [0]
from(x1) = [1 0] x1 + [0]
[1 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 1] [0 1] [0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)}
Weak Trs:
{ mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 1] [0 1] [1]
true() = [2]
[2]
mark(x1) = [0 1] x1 + [1]
[0 1] [0]
false() = [1]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 1] x2 + [0 1] x3 + [0]
[0 1] [0 1] [0 1] [2]
a__add(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 1] [0 1] [1]
0() = [2]
[2]
s(x1) = [0 0] x1 + [0]
[0 0] [0]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [3]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [3]
a__from(x1) = [0 0] x1 + [0]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 1] [0 1] [0 1] [2]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)}
Weak Trs:
{ mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {mark(add(X1, X2)) -> a__add(mark(X1), X2)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 1] x2 + [1]
[0 1] [0 1] [1]
true() = [0]
[0]
mark(x1) = [0 1] x1 + [0]
[0 1] [0]
false() = [0]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 1] x2 + [0 1] x3 + [0]
[0 1] [0 1] [0 1] [0]
a__add(x1, x2) = [1 0] x1 + [0 1] x2 + [1]
[0 1] [0 1] [2]
0() = [0]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [0]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [2]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
a__from(x1) = [0 0] x1 + [0]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 1] [0 1] [0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {mark(and(X1, X2)) -> a__and(mark(X1), X2)}
Weak Trs:
{ mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(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: {mark(and(X1, X2)) -> a__and(mark(X1), X2)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
Uargs(if) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a__and(x1, x2) = [1 0] x1 + [0 1] x2 + [1]
[0 1] [0 1] [2]
true() = [0]
[0]
mark(x1) = [0 1] x1 + [0]
[0 1] [0]
false() = [0]
[0]
a__if(x1, x2, x3) = [1 0] x1 + [0 1] x2 + [0 1] x3 + [0]
[0 1] [0 1] [0 1] [0]
a__add(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
[0 1] [0 1] [0]
0() = [0]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [0]
add(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
a__first(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
nil() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
first(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [0]
a__from(x1) = [0 0] x1 + [0]
[0 0] [0]
from(x1) = [0 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 1] [0 1] [2]
if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 1] [0 1] [0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Weak Trs:
{ mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, mark(cons(X1, X2)) -> cons(X1, X2)
, a__and(true(), X) -> mark(X)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(0(), X) -> mark(X)
, a__from(X) -> from(X)
, a__and(false(), Y) -> false()
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, mark(from(X)) -> a__from(X)
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, a__and(X1, X2) -> and(X1, X2)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__add(X1, X2) -> add(X1, X2)
, a__first(X1, X2) -> first(X1, X2)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
Hurray, we answered YES(?,O(n^1))