We consider the following Problem:
Strict Trs:
{ active(f(f(a()))) -> mark(f(g(f(a()))))
, active(f(X)) -> f(active(X))
, f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(a()) -> ok(a())
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(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(f(a()))) -> mark(f(g(f(a()))))
, active(f(X)) -> f(active(X))
, f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(a()) -> ok(a())
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(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: {proper(a()) -> ok(a())}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {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) = [0 0] x1 + [1]
[1 0] [1]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
a() = [0]
[1]
mark(x1) = [1 0] x1 + [1]
[0 0] [1]
g(x1) = [1 0] x1 + [0]
[0 0] [1]
proper(x1) = [0 0] x1 + [1]
[0 1] [0]
ok(x1) = [1 0] x1 + [0]
[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(f(a()))) -> mark(f(g(f(a()))))
, active(f(X)) -> f(active(X))
, f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))}
Weak Trs: {proper(a()) -> ok(a())}
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(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) = [0 0] x1 + [1]
[0 0] [1]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
a() = [0]
[0]
mark(x1) = [1 0] x1 + [1]
[0 0] [3]
g(x1) = [1 0] x1 + [0]
[0 0] [1]
proper(x1) = [0 0] x1 + [1]
[0 0] [1]
ok(x1) = [1 0] x1 + [0]
[0 0] [1]
top(x1) = [1 3] x1 + [1]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(f(a()))) -> mark(f(g(f(a()))))
, active(f(X)) -> f(active(X))
, f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, top(ok(X)) -> top(active(X))}
Weak Trs:
{ top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
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(f(a()))) -> mark(f(g(f(a()))))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {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) = [0 0] x1 + [1]
[0 0] [3]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
a() = [0]
[0]
mark(x1) = [1 0] x1 + [0]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 0] [1]
proper(x1) = [0 0] x1 + [0]
[0 0] [0]
ok(x1) = [1 0] x1 + [0]
[0 0] [0]
top(x1) = [1 1] 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))
, f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, top(ok(X)) -> top(active(X))}
Weak Trs:
{ active(f(f(a()))) -> mark(f(g(f(a()))))
, top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
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)) -> f(active(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(g) = {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) = [0 2] x1 + [2]
[0 1] [0]
f(x1) = [1 0] x1 + [0]
[0 1] [1]
a() = [0]
[0]
mark(x1) = [1 0] x1 + [0]
[0 0] [0]
g(x1) = [1 1] x1 + [3]
[0 0] [0]
proper(x1) = [0 0] x1 + [0]
[0 0] [0]
ok(x1) = [1 2] x1 + [0]
[0 0] [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:
{ f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))
, top(ok(X)) -> top(active(X))}
Weak Trs:
{ active(f(X)) -> f(active(X))
, active(f(f(a()))) -> mark(f(g(f(a()))))
, top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
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(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) = [0 0] x1 + [0]
[0 0] [0]
f(x1) = [1 0] x1 + [0]
[0 0] [0]
a() = [0]
[1]
mark(x1) = [1 0] x1 + [0]
[0 1] [0]
g(x1) = [1 0] x1 + [0]
[0 1] [0]
proper(x1) = [0 0] x1 + [0]
[0 1] [0]
ok(x1) = [1 0] x1 + [0]
[0 0] [1]
top(x1) = [1 1] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ f(mark(X)) -> mark(f(X))
, proper(f(X)) -> f(proper(X))
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))}
Weak Trs:
{ top(ok(X)) -> top(active(X))
, active(f(X)) -> f(active(X))
, active(f(f(a()))) -> mark(f(g(f(a()))))
, top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
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(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) = [0 1] x1 + [0]
[0 1] [0]
f(x1) = [1 0] x1 + [0]
[0 1] [1]
a() = [0]
[0]
mark(x1) = [1 1] x1 + [0]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 1] [0]
proper(x1) = [0 1] x1 + [0]
[0 1] [0]
ok(x1) = [1 1] x1 + [0]
[0 0] [0]
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:
{ f(mark(X)) -> mark(f(X))
, proper(g(X)) -> g(proper(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))}
Weak Trs:
{ proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> f(active(X))
, active(f(f(a()))) -> mark(f(g(f(a()))))
, top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
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(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) = [0 3] x1 + [0]
[0 1] [0]
f(x1) = [1 0] x1 + [0]
[0 1] [1]
a() = [0]
[0]
mark(x1) = [1 1] x1 + [1]
[0 0] [0]
g(x1) = [1 0] x1 + [0]
[0 1] [3]
proper(x1) = [0 1] x1 + [1]
[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:
{ f(mark(X)) -> mark(f(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))}
Weak Trs:
{ proper(g(X)) -> g(proper(X))
, proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> f(active(X))
, active(f(f(a()))) -> mark(f(g(f(a()))))
, top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ f(mark(X)) -> mark(f(X))
, f(ok(X)) -> ok(f(X))
, g(ok(X)) -> ok(g(X))}
Weak Trs:
{ proper(g(X)) -> g(proper(X))
, proper(f(X)) -> f(proper(X))
, top(ok(X)) -> top(active(X))
, active(f(X)) -> f(active(X))
, active(f(f(a()))) -> mark(f(g(f(a()))))
, top(mark(X)) -> top(proper(X))
, proper(a()) -> ok(a())}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 2.
The enriched problem is compatible with the following automaton:
{ active_0(2) -> 1
, active_1(2) -> 5
, active_1(3) -> 4
, active_2(7) -> 10
, active_2(8) -> 9
, f_0(2) -> 1
, f_1(2) -> 3
, f_1(5) -> 4
, f_1(6) -> 4
, f_2(7) -> 8
, f_2(10) -> 9
, a_0() -> 2
, a_1() -> 7
, mark_0(2) -> 2
, mark_1(3) -> 1
, mark_1(3) -> 3
, g_0(2) -> 1
, g_1(2) -> 3
, g_1(6) -> 4
, g_2(7) -> 8
, proper_0(2) -> 1
, proper_1(2) -> 6
, proper_1(3) -> 4
, ok_0(2) -> 1
, ok_0(2) -> 2
, ok_1(3) -> 1
, ok_1(3) -> 3
, ok_1(7) -> 6
, ok_2(8) -> 4
, top_0(1) -> 1
, top_0(2) -> 1
, top_1(4) -> 1
, top_2(9) -> 1}
Hurray, we answered YES(?,O(n^1))