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