Tool CaT
stdout:
YES(?,O(n^1))
Problem:
active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
mat(f(x),f(y())) -> f(mat(x,y()))
chk(no(c())) -> active(c())
mat(f(x),c()) -> no(c())
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))),x)))
Proof:
Bounds Processor:
bound: 2
enrichment: match
automaton:
final states: {10,9,8,7,6}
transitions:
tp1(27) -> 10*
chk1(26) -> 27*
mat1(25,1) -> 26*
mat1(25,3) -> 26*
mat1(25,5) -> 26*
mat1(25,2) -> 26*
mat1(25,4) -> 26*
f1(20) -> 21*
f1(15) -> 16*
f1(5) -> 13*
f1(22) -> 23*
f1(17) -> 18*
f1(2) -> 13*
f1(24) -> 25*
f1(19) -> 20*
f1(4) -> 13*
f1(21) -> 22*
f1(16) -> 17*
f1(1) -> 13*
f1(23) -> 24*
f1(18) -> 19*
f1(3) -> 13*
X1() -> 15*
mark1(13) -> 13,9
no1(12) -> 26*
no1(13) -> 13,9
active1(12) -> 7*
c1() -> 12*
active2(41) -> 27*
c2() -> 41*
active0(5) -> 6*
active0(2) -> 6*
active0(4) -> 6*
active0(1) -> 6*
active0(3) -> 6*
f0(5) -> 9*
f0(2) -> 9*
f0(4) -> 9*
f0(1) -> 9*
f0(3) -> 9*
mark0(5) -> 1*
mark0(2) -> 1*
mark0(4) -> 1*
mark0(1) -> 1*
mark0(3) -> 1*
chk0(5) -> 7*
chk0(2) -> 7*
chk0(4) -> 7*
chk0(1) -> 7*
chk0(3) -> 7*
no0(5) -> 2*
no0(2) -> 2*
no0(4) -> 2*
no0(1) -> 2*
no0(3) -> 2*
mat0(3,1) -> 8*
mat0(3,3) -> 8*
mat0(3,5) -> 8*
mat0(4,2) -> 8*
mat0(4,4) -> 8*
mat0(5,1) -> 8*
mat0(5,3) -> 8*
mat0(5,5) -> 8*
mat0(1,2) -> 8*
mat0(1,4) -> 8*
mat0(2,1) -> 8*
mat0(2,3) -> 8*
mat0(2,5) -> 8*
mat0(3,2) -> 8*
mat0(3,4) -> 8*
mat0(4,1) -> 8*
mat0(4,3) -> 8*
mat0(4,5) -> 8*
mat0(5,2) -> 8*
mat0(5,4) -> 8*
mat0(1,1) -> 8*
mat0(1,3) -> 8*
mat0(1,5) -> 8*
mat0(2,2) -> 8*
mat0(2,4) -> 8*
X0() -> 3*
y0() -> 4*
c0() -> 5*
tp0(5) -> 10*
tp0(2) -> 10*
tp0(4) -> 10*
tp0(1) -> 10*
tp0(3) -> 10*
problem:
QedTool IRC1
stdout:
YES(?,O(n^1))
Tool IRC2
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ active(f(x)) -> mark(f(f(x)))
, chk(no(f(x))) ->
f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))
, mat(f(x), f(y())) -> f(mat(x, y()))
, chk(no(c())) -> active(c())
, mat(f(x), c()) -> no(c())
, f(active(x)) -> active(f(x))
, f(no(x)) -> no(f(x))
, f(mark(x)) -> mark(f(x))
, tp(mark(x)) ->
tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))}
Proof Output:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:
Details:
--------
'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
'Bounds with minimal-enrichment and initial automaton 'match''
--------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ active(f(x)) -> mark(f(f(x)))
, chk(no(f(x))) ->
f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))
, mat(f(x), f(y())) -> f(mat(x, y()))
, chk(no(c())) -> active(c())
, mat(f(x), c()) -> no(c())
, f(active(x)) -> active(f(x))
, f(no(x)) -> no(f(x))
, f(mark(x)) -> mark(f(x))
, tp(mark(x)) ->
tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))}
Proof Output:
The problem is match-bounded by 2.
The enriched problem is compatible with the following automaton:
{ active_0(2) -> 1
, active_1(3) -> 1
, active_2(18) -> 5
, f_0(2) -> 1
, f_1(2) -> 4
, f_1(8) -> 7
, f_1(9) -> 8
, f_1(10) -> 9
, f_1(11) -> 10
, f_1(12) -> 11
, f_1(13) -> 12
, f_1(14) -> 13
, f_1(15) -> 14
, f_1(16) -> 15
, f_1(17) -> 16
, mark_0(2) -> 2
, mark_1(4) -> 1
, mark_1(4) -> 4
, chk_0(2) -> 1
, chk_1(6) -> 5
, no_0(2) -> 2
, no_1(3) -> 6
, no_1(4) -> 1
, no_1(4) -> 4
, mat_0(2, 2) -> 1
, mat_1(7, 2) -> 6
, X_0() -> 2
, X_1() -> 17
, y_0() -> 2
, c_0() -> 2
, c_1() -> 3
, c_2() -> 18
, tp_0(2) -> 1
, tp_1(5) -> 1}Tool RC1
stdout:
YES(?,O(n^1))
Tool RC2
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ active(f(x)) -> mark(f(f(x)))
, chk(no(f(x))) ->
f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))
, mat(f(x), f(y())) -> f(mat(x, y()))
, chk(no(c())) -> active(c())
, mat(f(x), c()) -> no(c())
, f(active(x)) -> active(f(x))
, f(no(x)) -> no(f(x))
, f(mark(x)) -> mark(f(x))
, tp(mark(x)) ->
tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))}
Proof Output:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:
Details:
--------
'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
'Bounds with minimal-enrichment and initial automaton 'match''
--------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ active(f(x)) -> mark(f(f(x)))
, chk(no(f(x))) ->
f(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))
, mat(f(x), f(y())) -> f(mat(x, y()))
, chk(no(c())) -> active(c())
, mat(f(x), c()) -> no(c())
, f(active(x)) -> active(f(x))
, f(no(x)) -> no(f(x))
, f(mark(x)) -> mark(f(x))
, tp(mark(x)) ->
tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X())))))))))), x)))}
Proof Output:
The problem is match-bounded by 2.
The enriched problem is compatible with the following automaton:
{ active_0(2) -> 1
, active_1(3) -> 1
, active_2(18) -> 5
, f_0(2) -> 1
, f_1(2) -> 4
, f_1(8) -> 7
, f_1(9) -> 8
, f_1(10) -> 9
, f_1(11) -> 10
, f_1(12) -> 11
, f_1(13) -> 12
, f_1(14) -> 13
, f_1(15) -> 14
, f_1(16) -> 15
, f_1(17) -> 16
, mark_0(2) -> 2
, mark_1(4) -> 1
, mark_1(4) -> 4
, chk_0(2) -> 1
, chk_1(6) -> 5
, no_0(2) -> 2
, no_1(3) -> 6
, no_1(4) -> 1
, no_1(4) -> 4
, mat_0(2, 2) -> 1
, mat_1(7, 2) -> 6
, X_0() -> 2
, X_1() -> 17
, y_0() -> 2
, c_0() -> 2
, c_1() -> 3
, c_2() -> 18
, tp_0(2) -> 1
, tp_1(5) -> 1}