details

property	value
status	complete
benchmark	Ex04.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n049.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.18617606163 seconds
cpu usage	0.193237102
max memory	9117696.0

stage attributes

key	value
output-size	2134
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 = round#(x) -> if#(x % 2 = 0, x, x + 1) minusNat#(true, I2, y) -> minus#(I2, round(y)) minusNat#(true, I2, y) -> round#(y) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(y)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = round#(x) -> if#(x % 2 = 0, x, x + 1) minusNat#(true, I2, y) -> minus#(I2, round(y)) minusNat#(true, I2, y) -> round#(y) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) minus#(I3, I4) -> minus#(I3, round(I4)) [I4 >= 0 && I3 = I4 + 1] minus#(I3, I4) -> round#(I4) [I4 >= 0 && I3 = I4 + 1] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(y)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) The dependency graph for this problem is: 0 -> 1 -> 5, 4, 3 2 -> 0 3 -> 4 -> 5, 4, 3 5 -> 0 Where: 0) round#(x) -> if#(x % 2 = 0, x, x + 1) 1) minusNat#(true, I2, y) -> minus#(I2, round(y)) 2) minusNat#(true, I2, y) -> round#(y) 3) minus#(I3, I4) -> minusNat#(I4 >= 0 && I3 = I4 + 1, I3, I4) 4) minus#(I3, I4) -> minus#(I3, round(I4)) [I4 >= 0 && I3 = I4 + 1] 5) minus#(I3, I4) -> round#(I4) [I4 >= 0 && I3 = I4 + 1] We have the following SCCs. { 4 } DP problem for innermost termination. P = minus#(I3, I4) -> minus#(I3, round(I4)) [I4 >= 0 && I3 = I4 + 1] R = if(false, u, v) -> v if(true, I0, I1) -> I0 round(x) -> if(x % 2 = 0, x, x + 1) minusNat(true, I2, y) -> minus(I2, round(y)) minus(I3, I4) -> minusNat(I4 >= 0 && I3 = I4 + 1, I3, I4) popout

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

job log

popout

actions

all output return to Integ TRS Inner 43557