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Integ TRS Inner 43557 pair #381740622
details
property
value
status
complete
benchmark
A08.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n094.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
2.53178405762 seconds
cpu usage
2.661386331
max memory
1.9116032E7
stage attributes
key
value
output-size
3817
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 = f#(true, x, y, z) -> f#(x > y && x > z, x, y, z + 1) f#(true, I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = f#(true, x, y, z) -> f_2#(x, y, z) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) f_1#(I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) The dependency graph for this problem is: 0 -> 5, 4, 3 1 -> 7, 6, 2 2 -> 3 -> 4 -> 7, 6, 2 5 -> 5, 3, 4 6 -> 7, 6, 2 7 -> 3, 4, 5 Where: 0) f#(true, x, y, z) -> f_2#(x, y, z) 1) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 2) f_1#(I0, I1, I2) -> f#(I0 > I1 && I0 > I2, I0, I1 + 1, I2) 3) f_2#(x, y, z) -> f#(x > y && x > z, x, y, z + 1) 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] 6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] 7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] We have the following SCCs. { 4, 5, 6, 7 } DP problem for innermost termination. P = f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 && I0 > I2] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) We use the reverse value criterion with the projection function NU: NU[f_1#(z1,z2,z3)] = z1 + -1 * z2 NU[f_2#(z1,z2,z3)] = z1 + -1 * z2 This gives the following inequalities: x > y && x > z ==> x + -1 * y >= x + -1 * y x > y && x > z ==> x + -1 * y >= x + -1 * y I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 I0 > I1 && I0 > I2 ==> I0 + -1 * I1 > I0 + -1 * (I1 + 1) with I0 + -1 * I1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2) The dependency graph for this problem is: 4 -> 5 -> 5, 4 Where: 4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y && x > z] 5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] We have the following SCCs. { 5 } DP problem for innermost termination. P = f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y && x > z] R = f(true, x, y, z) -> f(x > y && x > z, x, y, z + 1) f(true, I0, I1, I2) -> f(I0 > I1 && I0 > I2, I0, I1 + 1, I2)
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