details

property	value
status	complete
benchmark	A07.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n083.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.16145 seconds
cpu usage	1.17887
user time	0.60508
system time	0.573787
max virtual memory	124744.0
max residence set size	8624.0

stage attributes

key	value
starexec-result	YES

output

1.10/1.16 YES 1.10/1.16 1.10/1.16 DP problem for innermost termination. 1.10/1.16 P = 1.10/1.16 g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 1.10/1.16 g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 f#(true, I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 1.10/1.16 R = 1.10/1.16 g(true, x, y, z) -> f(x > z, x, y, z + 1) 1.10/1.16 g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 1.10/1.16 1.10/1.16 This problem is converted using chaining, where edges between chained DPs are removed. 1.10/1.16 1.10/1.16 DP problem for innermost termination. 1.10/1.16 P = 1.10/1.16 g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 1.10/1.16 g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 1.10/1.16 f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 1.10/1.16 f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4] 1.10/1.16 f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5] 1.10/1.16 f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4] 1.10/1.16 f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] 1.10/1.16 g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2] 1.10/1.16 g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z] 1.10/1.16 R = 1.10/1.16 g(true, x, y, z) -> f(x > z, x, y, z + 1) 1.10/1.16 g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 1.10/1.16 1.10/1.16 The dependency graph for this problem is: 1.10/1.16 0 -> 1.10/1.16 1 -> 1.10/1.16 2 -> 7, 6, 5, 4, 3 1.10/1.16 3 -> 1.10/1.16 4 -> 1.10/1.16 5 -> 7, 6, 5, 3, 4 1.10/1.16 6 -> 1.10/1.16 7 -> 7, 3, 4, 5, 6 1.10/1.16 8 -> 3, 4, 5, 6, 7 1.10/1.16 9 -> 3, 4, 5, 6, 7 1.10/1.16 Where: 1.10/1.16 0) g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 1.10/1.16 1) g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 2) f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 1.10/1.16 3) f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 1.10/1.16 4) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4] 1.10/1.16 5) f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5] 1.10/1.16 6) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4] 1.10/1.16 7) f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] 1.10/1.16 8) g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2] 1.10/1.16 9) g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z] 1.10/1.16 1.10/1.16 We have the following SCCs. 1.10/1.16 { 5, 7 } 1.10/1.16 1.10/1.16 DP problem for innermost termination. 1.10/1.16 P = 1.10/1.16 f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5] 1.10/1.16 f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] 1.10/1.16 R = 1.10/1.16 g(true, x, y, z) -> f(x > z, x, y, z + 1) 1.10/1.16 g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 1.10/1.16 1.10/1.16 We use the reverse value criterion with the projection function NU: 1.10/1.16 NU[f_1#(z1,z2,z3)] = z1 + -1 * z2 1.10/1.16 1.10/1.16 This gives the following inequalities: 1.10/1.16 I3 > I4/\I3 > I5 ==> I3 + -1 * I4 > I3 + -1 * (I4 + 1) with I3 + -1 * I4 >= 0 1.10/1.16 I3 > I4/\I3 > I5 ==> I3 + -1 * I4 >= I3 + -1 * I4 1.10/1.16 1.10/1.16 We remove all the strictly oriented dependency pairs. 1.10/1.16 1.10/1.16 DP problem for innermost termination. 1.10/1.16 P = 1.10/1.16 f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5] 1.10/1.16 R = 1.10/1.16 g(true, x, y, z) -> f(x > z, x, y, z + 1) 1.10/1.16 g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 1.10/1.16 f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 1.10/1.16 1.10/1.16 We use the reverse value criterion with the projection function NU: 1.10/1.16 NU[f_1#(z1,z2,z3)] = z1 + -1 * z3 1.10/1.16 1.10/1.16 This gives the following inequalities: 1.10/1.16 I3 > I4/\I3 > I5 ==> I3 + -1 * I5 > I3 + -1 * (I5 + 1) with I3 + -1 * I5 >= 0 1.10/1.16 1.10/1.16 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 1.10/4.14 EOF popout

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all output return to Integer_TRS_Innermost 2019-03-21 04.53