We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, active(if(true(), X, Y)) -> mark(X)
, active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(c()) -> active(c())
, mark(true()) -> active(true())
, mark(false()) -> active(false())
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, active(if(true(), X, Y)) -> mark(X)
, active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(c()) -> active(c())
, mark(true()) -> active(true())
, mark(false()) -> active(false())
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(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:
{ f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(X1, X2, X3)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
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]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
mark(x1) = [1 0] x1 + [1]
[1 0] [1]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 1] [0 0] [3]
c() = [0]
[0]
true() = [0]
[0]
false() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, active(if(true(), X, Y)) -> mark(X)
, active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(c()) -> active(c())
, mark(true()) -> active(true())
, mark(false()) -> active(false())
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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:
{ if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
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]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
mark(x1) = [1 0] x1 + [1]
[1 0] [1]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
c() = [0]
[0]
true() = [0]
[0]
false() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, active(if(true(), X, Y)) -> mark(X)
, active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(c()) -> active(c())
, mark(true()) -> active(true())
, mark(false()) -> active(false())
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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(if(true(), X, Y)) -> mark(X)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
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]
f(x1) = [1 0] x1 + [0]
[0 0] [0]
mark(x1) = [1 0] x1 + [1]
[1 0] [1]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
c() = [0]
[0]
true() = [2]
[0]
false() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(c()) -> active(c())
, mark(true()) -> active(true())
, mark(false()) -> active(false())
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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:
{ mark(c()) -> active(c())
, mark(false()) -> active(false())}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[1 0] [0]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
mark(x1) = [1 0] x1 + [3]
[0 0] [0]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
c() = [0]
[0]
true() = [2]
[0]
false() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(true()) -> active(true())
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ mark(c()) -> active(c())
, mark(false()) -> active(false())
, active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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(X)) -> mark(if(X, c(), f(true())))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[1 0] [0]
f(x1) = [1 0] x1 + [2]
[0 0] [1]
mark(x1) = [1 0] x1 + [1]
[1 0] [0]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
c() = [0]
[0]
true() = [0]
[0]
false() = [1]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, mark(true()) -> active(true())
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ active(f(X)) -> mark(if(X, c(), f(true())))
, mark(c()) -> active(c())
, mark(false()) -> active(false())
, active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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: {mark(true()) -> active(true())}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
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) = [1 0] x1 + [2]
[0 0] [1]
mark(x1) = [1 0] x1 + [2]
[0 0] [1]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [1 0] [0 0] [1]
c() = [0]
[0]
true() = [1]
[0]
false() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ active(if(false(), X, Y)) -> mark(Y)
, mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ mark(true()) -> active(true())
, active(f(X)) -> mark(if(X, c(), f(true())))
, mark(c()) -> active(c())
, mark(false()) -> active(false())
, active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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(if(false(), X, Y)) -> mark(Y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [0]
[1 0] [1]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
mark(x1) = [1 0] x1 + [0]
[1 0] [1]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
[1 0] [0 0] [0 0] [1]
c() = [0]
[0]
true() = [0]
[0]
false() = [2]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))
, if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Weak Trs:
{ active(if(false(), X, Y)) -> mark(Y)
, mark(true()) -> active(true())
, active(f(X)) -> mark(if(X, c(), f(true())))
, mark(c()) -> active(c())
, mark(false()) -> active(false())
, active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(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:
{ if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(active) = {1}, Uargs(f) = {1}, Uargs(mark) = {1},
Uargs(if) = {1, 2}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
active(x1) = [1 0] x1 + [1]
[1 0] [0]
f(x1) = [1 0] x1 + [0]
[0 0] [1]
mark(x1) = [1 0] x1 + [1]
[1 0] [0]
if(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
[0 0] [0 0] [1 0] [0]
c() = [0]
[0]
true() = [0]
[0]
false() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))}
Weak Trs:
{ if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)
, active(if(false(), X, Y)) -> mark(Y)
, mark(true()) -> active(true())
, active(f(X)) -> mark(if(X, c(), f(true())))
, mark(c()) -> active(c())
, mark(false()) -> active(false())
, active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(X1, X2, X3)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ mark(f(X)) -> active(f(mark(X)))
, mark(if(X1, X2, X3)) -> active(if(mark(X1), mark(X2), X3))}
Weak Trs:
{ if(X1, X2, mark(X3)) -> if(X1, X2, X3)
, if(X1, X2, active(X3)) -> if(X1, X2, X3)
, active(if(false(), X, Y)) -> mark(Y)
, mark(true()) -> active(true())
, active(f(X)) -> mark(if(X, c(), f(true())))
, mark(c()) -> active(c())
, mark(false()) -> active(false())
, active(if(true(), X, Y)) -> mark(X)
, if(X1, mark(X2), X3) -> if(X1, X2, X3)
, if(X1, active(X2), X3) -> if(X1, X2, X3)
, f(mark(X)) -> f(X)
, f(active(X)) -> f(X)
, if(mark(X1), X2, X3) -> if(X1, X2, X3)
, if(active(X1), X2, X3) -> if(X1, X2, X3)}
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
, f_0(2) -> 1
, mark_0(2) -> 1
, if_0(2, 2, 2) -> 1
, c_0() -> 2
, true_0() -> 2
, false_0() -> 2}
Hurray, we answered YES(?,O(n^1))