details

property	value
status	complete
benchmark	mult.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)	0.155318975449 seconds
cpu usage	0.126596525
max memory	8495104.0

stage attributes

key	value
output-size	1169
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = mult#(x, y) -> mult#(0 - x, y) [0 > x] mult#(I0, I1) -> mult#(I0 - 1, I1) [I0 > 0] R = mult(x, y) -> 0 - mult(0 - x, y) [0 > x] mult(I0, I1) -> mult(I0 - 1, I1) + I1 [I0 > 0] mult(0, I2) -> 0 The dependency graph for this problem is: 0 -> 1 1 -> 1 Where: 0) mult#(x, y) -> mult#(0 - x, y) [0 > x] 1) mult#(I0, I1) -> mult#(I0 - 1, I1) [I0 > 0] We have the following SCCs. { 1 } DP problem for innermost termination. P = mult#(I0, I1) -> mult#(I0 - 1, I1) [I0 > 0] R = mult(x, y) -> 0 - mult(0 - x, y) [0 > x] mult(I0, I1) -> mult(I0 - 1, I1) + I1 [I0 > 0] mult(0, I2) -> 0 We use the reverse value criterion with the projection function NU: NU[mult#(z1,z2)] = z1 This gives the following inequalities: I0 > 0 ==> I0 > I0 - 1 with I0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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

job log

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actions

all output return to Integer TRS Innermost