details

property	value
status	complete
benchmark	A06.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.77332 seconds
cpu usage	1.80689
user time	0.953612
system time	0.853279
max virtual memory	242940.0
max residence set size	8728.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f#(true, x, y, z) -> f#(x > y + z, x, y, z + 1) f#(true, I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) R = f(true, x, y, z) -> f(x > y + z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = f#(true, x, y, z) -> f_2#(x, y, z) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) f_1#(I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) f_2#(x, y, z) -> f#(x > y + z, x, y, z + 1) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z] f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2] f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2] R = f(true, x, y, z) -> f(x > y + z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) The dependency graph for this problem is: 0 -> 5, 4, 3 1 -> 7, 6, 2 2 -> 3 -> 4 -> 7, 6, 2 5 -> 5, 3, 4 6 -> 7, 6, 2 7 -> 3, 4, 5 Where: 0) f#(true, x, y, z) -> f_2#(x, y, z) 1) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 2) f_1#(I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) 3) f_2#(x, y, z) -> f#(x > y + z, x, y, z + 1) 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z] 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z] 6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2] 7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2] We have the following SCCs. { 4, 5, 6, 7 } DP problem for innermost termination. P = f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z] f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2] f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2] R = f(true, x, y, z) -> f(x > y + z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) We use the reverse value criterion with the projection function NU: NU[f_1#(z1,z2,z3)] = z1 + -1 * (z2 + 1 + z3) NU[f_2#(z1,z2,z3)] = z1 + -1 * (z2 + (z3 + 1)) This gives the following inequalities: x > y + z ==> x + -1 * (y + (z + 1)) > x + -1 * (y + 1 + (z + 1)) with x + -1 * (y + (z + 1)) >= 0 x > y + z ==> x + -1 * (y + (z + 1)) > x + -1 * (y + (z + 1 + 1)) with x + -1 * (y + (z + 1)) >= 0 I0 > I1 + I2 ==> I0 + -1 * (I1 + 1 + I2) > I0 + -1 * (I1 + 1 + 1 + I2) with I0 + -1 * (I1 + 1 + I2) >= 0 I0 > I1 + I2 ==> I0 + -1 * (I1 + 1 + I2) > I0 + -1 * (I1 + 1 + (I2 + 1)) with I0 + -1 * (I1 + 1 + I2) >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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job log

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actions

all output return to Integer TRS Innermost