We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(g(h(f(X))))
, active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(g(h(f(X))))
, active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {top(mark(X)) -> top(proper(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {1}, Uargs(h) = {1}, Uargs(proper) = {},
Uargs(ok) = {1}, Uargs(top) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[0 1] [0]
f(x1) = [1 1] x1 + [0]
[0 0] [0]
mark(x1) = [1 0] x1 + [1]
[0 1] [1]
g(x1) = [1 0] x1 + [0]
[0 0] [0]
h(x1) = [1 0] x1 + [0]
[0 0] [1]
proper(x1) = [0 0] x1 + [0]
[0 0] [1]
ok(x1) = [1 0] x1 + [0]
[0 1] [0]
top(x1) = [1 0] x1 + [0]
[0 1] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(g(h(f(X))))
, active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))
, top(ok(X)) -> top(active(X))}
Weak Trs: {top(mark(X)) -> top(proper(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(f(X)) -> mark(g(h(f(X))))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {1}, Uargs(h) = {1}, Uargs(proper) = {},
Uargs(ok) = {1}, Uargs(top) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[0 1] [1]
f(x1) = [1 1] x1 + [0]
[0 0] [0]
mark(x1) = [1 0] x1 + [0]
[0 1] [1]
g(x1) = [1 0] x1 + [0]
[0 0] [0]
h(x1) = [1 0] x1 + [0]
[0 0] [0]
proper(x1) = [0 0] x1 + [0]
[0 0] [0]
ok(x1) = [1 0] x1 + [0]
[0 1] [0]
top(x1) = [1 0] x1 + [1]
[0 1] [3]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))
, top(ok(X)) -> top(active(X))}
Weak Trs:
{ active(f(X)) -> mark(g(h(f(X))))
, top(mark(X)) -> top(proper(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {top(ok(X)) -> top(active(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {1}, Uargs(h) = {1}, Uargs(proper) = {},
Uargs(ok) = {1}, Uargs(top) = {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]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
mark(x1) = [1 0] x1 + [0]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 0] [1]
h(x1) = [1 0] x1 + [0]
[0 0] [1]
proper(x1) = [0 0] x1 + [0]
[0 1] [0]
ok(x1) = [1 0] x1 + [2]
[0 0] [1]
top(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(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))}
Weak Trs:
{ top(ok(X)) -> top(active(X))
, active(f(X)) -> mark(g(h(f(X))))
, top(mark(X)) -> top(proper(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {proper(f(X)) -> f(proper(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {1}, Uargs(h) = {1}, Uargs(proper) = {},
Uargs(ok) = {1}, Uargs(top) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 3] x1 + [1]
[0 0] [0]
f(x1) = [1 0] x1 + [0]
[0 1] [1]
mark(x1) = [1 1] x1 + [0]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 1] [0]
h(x1) = [1 0] x1 + [0]
[0 1] [0]
proper(x1) = [0 1] x1 + [0]
[0 1] [0]
ok(x1) = [1 3] x1 + [1]
[0 0] [0]
top(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:
{ active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(g(X)) -> g(proper(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))}
Weak Trs:
{ proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> mark(g(h(f(X))))
, top(mark(X)) -> top(proper(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {proper(g(X)) -> g(proper(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {1}, Uargs(h) = {1}, Uargs(proper) = {},
Uargs(ok) = {1}, Uargs(top) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 1] x1 + [1]
[0 0] [0]
f(x1) = [1 0] x1 + [0]
[0 1] [0]
mark(x1) = [1 1] x1 + [0]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 1] [1]
h(x1) = [1 0] x1 + [0]
[0 1] [0]
proper(x1) = [0 1] x1 + [0]
[0 1] [0]
ok(x1) = [1 1] x1 + [1]
[0 0] [0]
top(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:
{ active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, proper(h(X)) -> h(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))}
Weak Trs:
{ proper(g(X)) -> g(proper(X))
, proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> mark(g(h(f(X))))
, top(mark(X)) -> top(proper(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {proper(h(X)) -> h(proper(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {1}, Uargs(h) = {1}, Uargs(proper) = {},
Uargs(ok) = {1}, Uargs(top) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 1] x1 + [1]
[0 0] [0]
f(x1) = [1 0] x1 + [0]
[0 1] [0]
mark(x1) = [1 1] x1 + [0]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 1] [0]
h(x1) = [1 0] x1 + [0]
[0 1] [1]
proper(x1) = [0 1] x1 + [0]
[0 1] [0]
ok(x1) = [1 1] x1 + [1]
[0 0] [0]
top(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:
{ active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))}
Weak Trs:
{ proper(h(X)) -> h(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> mark(g(h(f(X))))
, top(mark(X)) -> top(proper(X))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> f(active(X))
, active(h(X)) -> h(active(X))
, f(mark(X)) -> mark(f(X))
, h(mark(X)) -> mark(h(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, h(ok(X)) -> ok(h(X))}
Weak Trs:
{ proper(h(X)) -> h(proper(X))
, proper(g(X)) -> g(proper(X))
, proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> mark(g(h(f(X))))
, top(mark(X)) -> top(proper(X))}
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:
{ active_0(3) -> 1
, active_0(7) -> 1
, f_0(3) -> 2
, f_0(7) -> 2
, f_1(3) -> 9
, f_1(7) -> 9
, mark_0(3) -> 3
, mark_0(7) -> 3
, mark_1(9) -> 2
, mark_1(9) -> 9
, mark_1(10) -> 5
, mark_1(10) -> 10
, g_0(3) -> 4
, g_0(7) -> 4
, g_1(3) -> 11
, g_1(7) -> 11
, h_0(3) -> 5
, h_0(7) -> 5
, h_1(3) -> 10
, h_1(7) -> 10
, proper_0(3) -> 6
, proper_0(7) -> 6
, ok_0(3) -> 7
, ok_0(7) -> 7
, ok_1(9) -> 2
, ok_1(9) -> 9
, ok_1(10) -> 5
, ok_1(10) -> 10
, ok_1(11) -> 4
, ok_1(11) -> 11
, top_0(1) -> 8
, top_0(3) -> 8
, top_0(6) -> 8
, top_0(7) -> 8}
Hurray, we answered YES(?,O(n^1))