details

property	value
status	complete
benchmark	poly4.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	1.10293 seconds
cpu usage	1.1079
user time	0.626654
system time	0.48125
max virtual memory	124360.0
max residence set size	8580.0

stage attributes

key	value
starexec-result	YES

output

1.08/1.10 YES 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 R = 1.08/1.10 eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 1.08/1.10 The dependency graph for this problem is: 1.08/1.10 0 -> 0 1.08/1.10 1 -> 1 1.08/1.10 2 -> 0, 2, 3 1.08/1.10 3 -> 1, 2, 3 1.08/1.10 Where: 1.08/1.10 0) eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 1) eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 2) eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 3) eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 1.08/1.10 We have the following SCCs. 1.08/1.10 { 2, 3 } 1.08/1.10 { 0 } 1.08/1.10 { 1 } 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 eval#(I0, I1, I2, I3, I4) -> eval#(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 R = 1.08/1.10 eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 1.08/1.10 We use the reverse value criterion with the projection function NU: 1.08/1.10 NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4 1.08/1.10 1.08/1.10 This gives the following inequalities: 1.08/1.10 I4 >= I1 && I3 < I0 ==> I0 + -1 * I3 > I0 + -1 * (I3 + 1) with I0 + -1 * I3 >= 0 1.08/1.10 1.08/1.10 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 eval#(an, bn, cn, i, j) -> eval#(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 R = 1.08/1.10 eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 1.08/1.10 We use the reverse value criterion with the projection function NU: 1.08/1.10 NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5 1.08/1.10 1.08/1.10 This gives the following inequalities: 1.08/1.10 j < bn && i >= an ==> bn + -1 * j > bn + -1 * (j + 1) with bn + -1 * j >= 0 1.08/1.10 1.08/1.10 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 eval#(I5, I6, I7, I8, I9) -> eval#(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 R = 1.08/1.10 eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 1.08/1.10 We use the reverse value criterion with the projection function NU: 1.08/1.10 NU[eval#(z1,z2,z3,z4,z5)] = z1 + -1 * z4 1.08/1.10 1.08/1.10 This gives the following inequalities: 1.08/1.10 I9 < I6 && I8 < I5 ==> I5 + -1 * I8 > I5 + -1 * (I8 + 1) with I5 + -1 * I8 >= 0 1.08/1.10 I14 < I11 && I13 < I10 ==> I10 + -1 * I13 >= I10 + -1 * I13 1.08/1.10 1.08/1.10 We remove all the strictly oriented dependency pairs. 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 eval#(I10, I11, I12, I13, I14) -> eval#(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 R = 1.08/1.10 eval(an, bn, cn, i, j) -> eval(an, bn, cn + 1, i, j + 1) [j < bn && i >= an] 1.08/1.10 eval(I0, I1, I2, I3, I4) -> eval(I0, I1, I2 + 1, I3 + 1, I4) [I4 >= I1 && I3 < I0] 1.08/1.10 eval(I5, I6, I7, I8, I9) -> eval(I5, I6, I7 + 1, I8 + 1, I9) [I9 < I6 && I8 < I5] 1.08/1.10 eval(I10, I11, I12, I13, I14) -> eval(I10, I11, I12 + 1, I13, I14 + 1) [I14 < I11 && I13 < I10] 1.08/1.10 1.08/1.10 We use the reverse value criterion with the projection function NU: 1.08/1.10 NU[eval#(z1,z2,z3,z4,z5)] = z2 + -1 * z5 1.08/1.10 1.08/1.10 This gives the following inequalities: 1.08/1.10 I14 < I11 && I13 < I10 ==> I11 + -1 * I14 > I11 + -1 * (I14 + 1) with I11 + -1 * I14 >= 0 1.08/1.10 1.08/1.10 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 1.08/1.10 EOF popout

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all output return to Integer_TRS_Innermost 2019-03-21 04.53