We consider the following Problem:
Strict Trs:
{ active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, 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(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, 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: {f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1, 2, 3}, Uargs(mark) = {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 0] [1]
f(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
a() = [0]
[0]
mark(x1) = [1 0] x1 + [1]
[1 0] [1]
b() = [0]
[0]
proper(x1) = [0 0] x1 + [1]
[0 0] [1]
ok(x1) = [1 0] x1 + [1]
[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(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))}
Weak Trs: {f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
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, 2, 3}, Uargs(mark) = {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 0] [1]
f(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
a() = [0]
[1]
mark(x1) = [1 0] x1 + [1]
[1 0] [1]
b() = [0]
[0]
proper(x1) = [0 0] x1 + [1]
[0 0] [0]
ok(x1) = [1 0] x1 + [3]
[0 1] [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:
{ active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, top(mark(X)) -> top(proper(X))}
Weak Trs:
{ top(ok(X)) -> top(active(X))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ active(b()) -> mark(a())
, top(mark(X)) -> top(proper(X))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1, 2, 3}, Uargs(mark) = {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] [1]
f(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 0] [0 0] [1]
a() = [0]
[0]
mark(x1) = [1 0] x1 + [1]
[1 0] [0]
b() = [2]
[0]
proper(x1) = [0 0] x1 + [0]
[0 0] [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(a(), X, X)) -> mark(f(X, b(), b()))
, active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())}
Weak Trs:
{ active(b()) -> mark(a())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
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(a(), X, X)) -> mark(f(X, b(), b()))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1, 2, 3}, Uargs(mark) = {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 + [2]
[0 0] [1]
f(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 1] [0 0] [0]
a() = [3]
[0]
mark(x1) = [1 0] x1 + [0]
[0 0] [1]
b() = [2]
[0]
proper(x1) = [0 0] x1 + [0]
[0 0] [1]
ok(x1) = [1 0] x1 + [2]
[0 0] [1]
top(x1) = [1 0] x1 + [3]
[0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())}
Weak Trs:
{ active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {proper(b()) -> ok(b())}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {}, Uargs(f) = {1, 2, 3}, Uargs(mark) = {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, x2, x3) = [1 1] x1 + [1 0] x2 + [1 2] x3 + [0]
[0 0] [0 0] [0 0] [0]
a() = [0]
[3]
mark(x1) = [1 0] x1 + [1]
[0 0] [0]
b() = [1]
[0]
proper(x1) = [1 0] x1 + [1]
[0 0] [0]
ok(x1) = [1 3] 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:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())}
Weak Trs:
{ proper(b()) -> ok(b())
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
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, 2, 3}, Uargs(mark) = {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, x2, x3) = [1 3] x1 + [1 0] x2 + [1 3] x3 + [0]
[0 0] [0 0] [0 0] [0]
a() = [0]
[3]
mark(x1) = [1 0] x1 + [1]
[0 0] [0]
b() = [1]
[0]
proper(x1) = [1 0] x1 + [1]
[0 1] [0]
ok(x1) = [1 0] x1 + [0]
[0 1] [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(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))}
Weak Trs:
{ proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))}
Weak Trs:
{ proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))}
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(4) -> 1
, active_0(5) -> 1
, active_0(7) -> 1
, f_0(3, 3, 3) -> 2
, f_0(3, 3, 4) -> 2
, f_0(3, 3, 5) -> 2
, f_0(3, 3, 7) -> 2
, f_0(3, 4, 3) -> 2
, f_0(3, 4, 4) -> 2
, f_0(3, 4, 5) -> 2
, f_0(3, 4, 7) -> 2
, f_0(3, 5, 3) -> 2
, f_0(3, 5, 4) -> 2
, f_0(3, 5, 5) -> 2
, f_0(3, 5, 7) -> 2
, f_0(3, 7, 3) -> 2
, f_0(3, 7, 4) -> 2
, f_0(3, 7, 5) -> 2
, f_0(3, 7, 7) -> 2
, f_0(4, 3, 3) -> 2
, f_0(4, 3, 4) -> 2
, f_0(4, 3, 5) -> 2
, f_0(4, 3, 7) -> 2
, f_0(4, 4, 3) -> 2
, f_0(4, 4, 4) -> 2
, f_0(4, 4, 5) -> 2
, f_0(4, 4, 7) -> 2
, f_0(4, 5, 3) -> 2
, f_0(4, 5, 4) -> 2
, f_0(4, 5, 5) -> 2
, f_0(4, 5, 7) -> 2
, f_0(4, 7, 3) -> 2
, f_0(4, 7, 4) -> 2
, f_0(4, 7, 5) -> 2
, f_0(4, 7, 7) -> 2
, f_0(5, 3, 3) -> 2
, f_0(5, 3, 4) -> 2
, f_0(5, 3, 5) -> 2
, f_0(5, 3, 7) -> 2
, f_0(5, 4, 3) -> 2
, f_0(5, 4, 4) -> 2
, f_0(5, 4, 5) -> 2
, f_0(5, 4, 7) -> 2
, f_0(5, 5, 3) -> 2
, f_0(5, 5, 4) -> 2
, f_0(5, 5, 5) -> 2
, f_0(5, 5, 7) -> 2
, f_0(5, 7, 3) -> 2
, f_0(5, 7, 4) -> 2
, f_0(5, 7, 5) -> 2
, f_0(5, 7, 7) -> 2
, f_0(7, 3, 3) -> 2
, f_0(7, 3, 4) -> 2
, f_0(7, 3, 5) -> 2
, f_0(7, 3, 7) -> 2
, f_0(7, 4, 3) -> 2
, f_0(7, 4, 4) -> 2
, f_0(7, 4, 5) -> 2
, f_0(7, 4, 7) -> 2
, f_0(7, 5, 3) -> 2
, f_0(7, 5, 4) -> 2
, f_0(7, 5, 5) -> 2
, f_0(7, 5, 7) -> 2
, f_0(7, 7, 3) -> 2
, f_0(7, 7, 4) -> 2
, f_0(7, 7, 5) -> 2
, f_0(7, 7, 7) -> 2
, f_1(3, 3, 3) -> 9
, f_1(3, 3, 4) -> 9
, f_1(3, 3, 5) -> 9
, f_1(3, 3, 7) -> 9
, f_1(3, 4, 3) -> 9
, f_1(3, 4, 4) -> 9
, f_1(3, 4, 5) -> 9
, f_1(3, 4, 7) -> 9
, f_1(3, 5, 3) -> 9
, f_1(3, 5, 4) -> 9
, f_1(3, 5, 5) -> 9
, f_1(3, 5, 7) -> 9
, f_1(3, 7, 3) -> 9
, f_1(3, 7, 4) -> 9
, f_1(3, 7, 5) -> 9
, f_1(3, 7, 7) -> 9
, f_1(4, 3, 3) -> 9
, f_1(4, 3, 4) -> 9
, f_1(4, 3, 5) -> 9
, f_1(4, 3, 7) -> 9
, f_1(4, 4, 3) -> 9
, f_1(4, 4, 4) -> 9
, f_1(4, 4, 5) -> 9
, f_1(4, 4, 7) -> 9
, f_1(4, 5, 3) -> 9
, f_1(4, 5, 4) -> 9
, f_1(4, 5, 5) -> 9
, f_1(4, 5, 7) -> 9
, f_1(4, 7, 3) -> 9
, f_1(4, 7, 4) -> 9
, f_1(4, 7, 5) -> 9
, f_1(4, 7, 7) -> 9
, f_1(5, 3, 3) -> 9
, f_1(5, 3, 4) -> 9
, f_1(5, 3, 5) -> 9
, f_1(5, 3, 7) -> 9
, f_1(5, 4, 3) -> 9
, f_1(5, 4, 4) -> 9
, f_1(5, 4, 5) -> 9
, f_1(5, 4, 7) -> 9
, f_1(5, 5, 3) -> 9
, f_1(5, 5, 4) -> 9
, f_1(5, 5, 5) -> 9
, f_1(5, 5, 7) -> 9
, f_1(5, 7, 3) -> 9
, f_1(5, 7, 4) -> 9
, f_1(5, 7, 5) -> 9
, f_1(5, 7, 7) -> 9
, f_1(7, 3, 3) -> 9
, f_1(7, 3, 4) -> 9
, f_1(7, 3, 5) -> 9
, f_1(7, 3, 7) -> 9
, f_1(7, 4, 3) -> 9
, f_1(7, 4, 4) -> 9
, f_1(7, 4, 5) -> 9
, f_1(7, 4, 7) -> 9
, f_1(7, 5, 3) -> 9
, f_1(7, 5, 4) -> 9
, f_1(7, 5, 5) -> 9
, f_1(7, 5, 7) -> 9
, f_1(7, 7, 3) -> 9
, f_1(7, 7, 4) -> 9
, f_1(7, 7, 5) -> 9
, f_1(7, 7, 7) -> 9
, a_0() -> 3
, mark_0(3) -> 1
, mark_0(3) -> 4
, mark_0(4) -> 4
, mark_0(5) -> 4
, mark_0(7) -> 4
, mark_1(9) -> 2
, mark_1(9) -> 9
, b_0() -> 5
, proper_0(3) -> 6
, proper_0(4) -> 6
, proper_0(5) -> 6
, proper_0(7) -> 6
, ok_0(2) -> 2
, ok_0(3) -> 6
, ok_0(3) -> 7
, ok_0(4) -> 7
, ok_0(5) -> 6
, ok_0(5) -> 7
, ok_0(7) -> 7
, ok_1(9) -> 9
, top_0(1) -> 8
, top_0(3) -> 8
, top_0(4) -> 8
, top_0(5) -> 8
, top_0(6) -> 8
, top_0(7) -> 8}
Hurray, we answered YES(?,O(n^1))