details

property	value
status	complete
benchmark	A08.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	3.00452 seconds
cpu usage	2.60288
user time	1.46289
system time	1.13999
max virtual memory	728156.0
max residence set size	8652.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f#(true, x, y, z) -> f#(x > y && x > z, x, y, z + 1) f#(true, I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > 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 && I0 > I2, I0, I1 + 1, I2) f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > 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 && I0 > I2, I0, I1 + 1, I2) 3) f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > 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 && x > z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > 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 NU[f_2#(z1,z2,z3)] = z1 + -1 * z2 This gives the following inequalities: x > y && x > z ==> x + -1 * y >= x + -1 * y x > y && x > z ==> x + -1 * y >= x + -1 * y I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) The dependency graph for this problem is: 4 -> 5 -> 5, 4 Where: 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] We have the following SCCs. { 5 } DP problem for innermost termination. P = f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) We use the reverse value criterion with the projection function NU: NU[f_2#(z1,z2,z3)] = z1 + -1 * z3 This gives the following inequalities: x > y && x > z ==> x + -1 * z > x + -1 * (z + 1) with x + -1 * z >= 0 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Integer TRS Innermost