details

property	value
status	complete
benchmark	csharp2.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n123.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	2.37042 seconds
cpu usage	2.41505
user time	1.23884
system time	1.17622
max virtual memory	729104.0
max residence set size	8612.0

stage attributes

key	value
starexec-result	YES

output

2.37/2.36 YES 2.37/2.36 2.37/2.36 DP problem for innermost termination. 2.37/2.36 P = 2.37/2.36 b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.36 b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.36 Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 2.37/2.36 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 2.37/2.36 R = 2.37/2.36 b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.36 b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.36 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 2.37/2.36 b10(I6, I7, I8) -> b14(I6, I7, I8) 2.37/2.36 2.37/2.36 This problem is converted using chaining, where edges between chained DPs are removed. 2.37/2.36 2.37/2.36 DP problem for innermost termination. 2.37/2.36 P = 2.37/2.36 b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.36 b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.36 Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 2.37/2.36 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 2.37/2.36 b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1] 2.37/2.37 R = 2.37/2.37 b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.37 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 2.37/2.37 b10(I6, I7, I8) -> b14(I6, I7, I8) 2.37/2.37 2.37/2.37 The dependency graph for this problem is: 2.37/2.37 0 -> 3 2.37/2.37 1 -> 2.37/2.37 2 -> 0 2.37/2.37 3 -> 4, 1 2.37/2.37 4 -> 0 2.37/2.37 Where: 2.37/2.37 0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 1) b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.37 2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 2.37/2.37 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 2.37/2.37 4) b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1] 2.37/2.37 2.37/2.37 We have the following SCCs. 2.37/2.37 { 0, 3, 4 } 2.37/2.37 2.37/2.37 DP problem for innermost termination. 2.37/2.37 P = 2.37/2.37 b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 2.37/2.37 b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1] 2.37/2.37 R = 2.37/2.37 b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.37 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 2.37/2.37 b10(I6, I7, I8) -> b14(I6, I7, I8) 2.37/2.37 2.37/2.37 We use the extended value criterion with the projection function NU: 2.37/2.37 NU[b14#(x0,x1,x2)] = x1 - 2 2.37/2.37 NU[b10#(x0,x1,x2)] = x1 - 2 2.37/2.37 NU[b15#(x0,x1,x2)] = -x0 + x1 - 2 2.37/2.37 2.37/2.37 This gives the following inequalities: 2.37/2.37 ==> -sv14_14 + sv23_37 - 2 >= (sv23_37 - sv14_14) - 2 2.37/2.37 ==> I7 - 2 >= I7 - 2 2.37/2.37 I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1 ==> I1 - 2 > -I0 + I1 - 2 with I1 - 2 >= 0 2.37/2.37 2.37/2.37 We remove all the strictly oriented dependency pairs. 2.37/2.37 2.37/2.37 DP problem for innermost termination. 2.37/2.37 P = 2.37/2.37 b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 b10#(I6, I7, I8) -> b14#(I6, I7, I8) 2.37/2.37 R = 2.37/2.37 b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 2.37/2.37 Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 2.37/2.37 b10(I6, I7, I8) -> b14(I6, I7, I8) 2.37/2.37 2.37/2.37 The dependency graph for this problem is: 2.37/2.37 0 -> 3 2.37/2.37 3 -> 2.37/2.37 Where: 2.37/2.37 0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 2.37/2.37 3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 2.37/2.37 2.37/2.37 We have the following SCCs. 2.37/2.37 2.37/5.34 EOF popout

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all output return to Integer_TRS_Innermost 2019-03-21 04.53