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Integ TRS Inner 43557 pair #381740678
details
property
value
status
complete
benchmark
16.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)
0.370840072632 seconds
cpu usage
0.385726626
max memory
9854976.0
stage attributes
key
value
output-size
2605
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 = eval#(x, y, z) -> eval#(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y] eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0] eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0] R = eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y] eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0] eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0] The dependency graph for this problem is: 0 -> 1 -> 1 2 -> 1, 2 Where: 0) eval#(x, y, z) -> eval#(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y] 1) eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0] 2) eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0] We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = eval#(I0, I1, I2) -> eval#(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0] R = eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y] eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0] eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z2 This gives the following inequalities: I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0 ==> I1 > I1 - 1 with I1 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = eval#(I3, I4, I5) -> eval#(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0] R = eval(x, y, z) -> eval(x, y, z) [x + y > z && z >= 0 && 0 >= x && 0 >= y] eval(I0, I1, I2) -> eval(I0, I1 - 1, I2) [I0 + I1 > I2 && I2 >= 0 && 0 >= I0 && I1 > 0] eval(I3, I4, I5) -> eval(I3 - 1, I4, I5) [I3 + I4 > I5 && I5 >= 0 && I3 > 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2,z3)] = z1 + z2 + -1 * z3 This gives the following inequalities: I3 + I4 > I5 && I5 >= 0 && I3 > 0 ==> I3 + I4 + -1 * I5 > I3 - 1 + I4 + -1 * I5 with I3 + I4 + -1 * I5 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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