We consider the following Problem:
Strict Trs:
{ active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, active(plus(N, s(M))) -> mark(s(plus(N, M)))
, mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(tt()) -> active(tt())
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(0()) -> active(0())
, mark(s(X)) -> active(s(mark(X)))
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, active(plus(N, s(M))) -> mark(s(plus(N, M)))
, mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(tt()) -> active(tt())
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(0()) -> active(0())
, mark(s(X)) -> active(s(mark(X)))
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(and) = {1}, Uargs(mark) = {1},
Uargs(plus) = {1, 2}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[1 0] [1]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
tt() = [0]
[0]
mark(x1) = [1 0] x1 + [1]
[0 0] [1]
plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, active(plus(N, s(M))) -> mark(s(plus(N, M)))
, mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(tt()) -> active(tt())
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(0()) -> active(0())
, mark(s(X)) -> active(s(mark(X)))}
Weak Trs:
{ and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ mark(tt()) -> active(tt())
, mark(0()) -> active(0())}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(and) = {1}, Uargs(mark) = {1},
Uargs(plus) = {1, 2}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[1 0] [1]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
tt() = [0]
[0]
mark(x1) = [1 0] x1 + [3]
[0 0] [1]
plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, active(plus(N, s(M))) -> mark(s(plus(N, M)))
, mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(s(X)) -> active(s(mark(X)))}
Weak Trs:
{ mark(tt()) -> active(tt())
, mark(0()) -> active(0())
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {active(plus(N, 0())) -> mark(N)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(and) = {1}, Uargs(mark) = {1},
Uargs(plus) = {1, 2}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[1 0] [1]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
tt() = [0]
[0]
mark(x1) = [1 0] x1 + [1]
[1 0] [1]
plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
0() = [3]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(and(tt(), X)) -> mark(X)
, active(plus(N, s(M))) -> mark(s(plus(N, M)))
, mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(s(X)) -> active(s(mark(X)))}
Weak Trs:
{ active(plus(N, 0())) -> mark(N)
, mark(tt()) -> active(tt())
, mark(0()) -> active(0())
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {active(and(tt(), X)) -> mark(X)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(and) = {1}, Uargs(mark) = {1},
Uargs(plus) = {1, 2}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [0]
[0 0] [0]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
tt() = [1]
[0]
mark(x1) = [1 0] x1 + [0]
[0 0] [0]
plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(plus(N, s(M))) -> mark(s(plus(N, M)))
, mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(s(X)) -> active(s(mark(X)))}
Weak Trs:
{ active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, mark(tt()) -> active(tt())
, mark(0()) -> active(0())
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {active(plus(N, s(M))) -> mark(s(plus(N, M)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(and) = {1}, Uargs(mark) = {1},
Uargs(plus) = {1, 2}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 2] x1 + [0]
[0 0] [1]
and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [2]
tt() = [0]
[0]
mark(x1) = [1 0] x1 + [0]
[0 0] [1]
plus(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [2]
0() = [0]
[0]
s(x1) = [1 0] x1 + [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)) -> active(and(mark(X1), X2))
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(s(X)) -> active(s(mark(X)))}
Weak Trs:
{ active(plus(N, s(M))) -> mark(s(plus(N, M)))
, active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, mark(tt()) -> active(tt())
, mark(0()) -> active(0())
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ mark(and(X1, X2)) -> active(and(mark(X1), X2))
, mark(plus(X1, X2)) -> active(plus(mark(X1), mark(X2)))
, mark(s(X)) -> active(s(mark(X)))}
Weak Trs:
{ active(plus(N, s(M))) -> mark(s(plus(N, M)))
, active(and(tt(), X)) -> mark(X)
, active(plus(N, 0())) -> mark(N)
, mark(tt()) -> active(tt())
, mark(0()) -> active(0())
, and(mark(X1), X2) -> and(X1, X2)
, and(X1, mark(X2)) -> and(X1, X2)
, and(active(X1), X2) -> and(X1, X2)
, and(X1, active(X2)) -> and(X1, X2)
, plus(mark(X1), X2) -> plus(X1, X2)
, plus(X1, mark(X2)) -> plus(X1, X2)
, plus(active(X1), X2) -> plus(X1, X2)
, plus(X1, active(X2)) -> plus(X1, X2)
, s(mark(X)) -> s(X)
, s(active(X)) -> s(X)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 0.
The enriched problem is compatible with the following automaton:
{ active_0(2) -> 1
, and_0(2, 2) -> 1
, tt_0() -> 2
, mark_0(2) -> 1
, plus_0(2, 2) -> 1
, 0_0() -> 2
, s_0(2) -> 1}
Hurray, we answered YES(?,O(n^1))