details

property	value
status	complete
benchmark	A17.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n020.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.365571022034 seconds
cpu usage	0.382327939
max memory	9994240.0

stage attributes

key	value
output-size	2074
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = cond#(false, x, y) -> log#(x, y * y) logNat#(true, I2, I3) -> cond#(I2 <= I3, I2, I3) log#(I4, I5) -> logNat#(I4 >= 0 && I5 >= 2, I4, I5) R = cond(false, x, y) -> 2 * log(x, y * y) cond(true, I0, I1) -> 1 logNat(true, I2, I3) -> cond(I2 <= I3, I2, I3) log(I4, I5) -> logNat(I4 >= 0 && I5 >= 2, I4, I5) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = cond#(false, x, y) -> log#(x, y * y) logNat#(true, I2, I3) -> cond#(I2 <= I3, I2, I3) log#(I4, I5) -> logNat#(I4 >= 0 && I5 >= 2, I4, I5) log#(I4, I5) -> cond#(I4 <= I5, I4, I5) [I4 >= 0 && I5 >= 2] log#(I4, I5) -> log#(I4, I5 * I5) [I4 >= 0 && I5 >= 2, not(I4 <= I5)] logNat#(true, I2, I3) -> log#(I2, I3 * I3) [not(I2 <= I3)] R = cond(false, x, y) -> 2 * log(x, y * y) cond(true, I0, I1) -> 1 logNat(true, I2, I3) -> cond(I2 <= I3, I2, I3) log(I4, I5) -> logNat(I4 >= 0 && I5 >= 2, I4, I5) The dependency graph for this problem is: 0 -> 4, 3, 2 1 -> 2 -> 3 -> 4 -> 4, 2, 3 5 -> 2, 3, 4 Where: 0) cond#(false, x, y) -> log#(x, y * y) 1) logNat#(true, I2, I3) -> cond#(I2 <= I3, I2, I3) 2) log#(I4, I5) -> logNat#(I4 >= 0 && I5 >= 2, I4, I5) 3) log#(I4, I5) -> cond#(I4 <= I5, I4, I5) [I4 >= 0 && I5 >= 2] 4) log#(I4, I5) -> log#(I4, I5 * I5) [I4 >= 0 && I5 >= 2, not(I4 <= I5)] 5) logNat#(true, I2, I3) -> log#(I2, I3 * I3) [not(I2 <= I3)] We have the following SCCs. { 4 } DP problem for innermost termination. P = log#(I4, I5) -> log#(I4, I5 * I5) [I4 >= 0 && I5 >= 2, not(I4 <= I5)] R = cond(false, x, y) -> 2 * log(x, y * y) cond(true, I0, I1) -> 1 logNat(true, I2, I3) -> cond(I2 <= I3, I2, I3) log(I4, I5) -> logNat(I4 >= 0 && I5 >= 2, I4, I5) popout

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

job log

popout

actions

all output return to Integ TRS Inner 43557