/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f10#(x1, x2, x3) -> f1#(x1, x2, x3) f9#(I0, I1, I2) -> f2#(I0, I1, I2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] f5#(I18, I19, I20) -> f6#(I18, I19, I20) [1 <= I19] f5#(I21, I22, I23) -> f6#(I21, I22, I23) [1 + I22 <= 0] f2#(I24, I25, I26) -> f5#(I24, I27, I26) [I27 = I27] f4#(I28, I29, I30) -> f2#(I28, I29, I30) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f1#(I40, I41, I42) -> f2#(I40, I41, I42) R = f10(x1, x2, x3) -> f1(x1, x2, x3) f9(I0, I1, I2) -> f2(I0, I1, I2) f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7] f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2] f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0] f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19] f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0] f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27] f4(I28, I29, I30) -> f2(I28, I29, I30) f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39] f1(I40, I41, I42) -> f2(I40, I41, I42) The dependency graph for this problem is: 0 -> 11 1 -> 5, 8, 10 2 -> 1 3 -> 2 4 -> 2 5 -> 3, 4 6 -> 7 -> 8 -> 6, 7 9 -> 5, 8, 10 10 -> 9 11 -> 5, 8, 10 Where: 0) f10#(x1, x2, x3) -> f1#(x1, x2, x3) 1) f9#(I0, I1, I2) -> f2#(I0, I1, I2) 2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] 3) f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] 4) f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] 5) f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] 6) f5#(I18, I19, I20) -> f6#(I18, I19, I20) [1 <= I19] 7) f5#(I21, I22, I23) -> f6#(I21, I22, I23) [1 + I22 <= 0] 8) f2#(I24, I25, I26) -> f5#(I24, I27, I26) [I27 = I27] 9) f4#(I28, I29, I30) -> f2#(I28, I29, I30) 10) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] 11) f1#(I40, I41, I42) -> f2#(I40, I41, I42) We have the following SCCs. { 1, 2, 3, 4, 5, 9, 10 } DP problem for innermost termination. P = f9#(I0, I1, I2) -> f2#(I0, I1, I2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] f4#(I28, I29, I30) -> f2#(I28, I29, I30) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] R = f10(x1, x2, x3) -> f1(x1, x2, x3) f9(I0, I1, I2) -> f2(I0, I1, I2) f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7] f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2] f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0] f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19] f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0] f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27] f4(I28, I29, I30) -> f2(I28, I29, I30) f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39] f1(I40, I41, I42) -> f2(I40, I41, I42) We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = z3 NU[f7#(z1,z2,z3)] = z3 NU[f8#(z1,z2,z3)] = z3 NU[f2#(z1,z2,z3)] = z3 NU[f9#(z1,z2,z3)] = z3 This gives the following inequalities: ==> I2 (>! \union =) I2 0 <= -1 + -1 + I5 ==> I5 >! -1 + I5 1 <= I7 ==> I8 (>! \union =) I8 1 + I10 <= 0 ==> I11 (>! \union =) I11 rnd2 = rnd2 ==> I14 (>! \union =) I14 ==> I30 (>! \union =) I30 0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34 ==> I33 (>! \union =) I33 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f9#(I0, I1, I2) -> f2#(I0, I1, I2) f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] f4#(I28, I29, I30) -> f2#(I28, I29, I30) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] R = f10(x1, x2, x3) -> f1(x1, x2, x3) f9(I0, I1, I2) -> f2(I0, I1, I2) f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7] f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2] f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0] f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19] f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0] f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27] f4(I28, I29, I30) -> f2(I28, I29, I30) f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39] f1(I40, I41, I42) -> f2(I40, I41, I42) The dependency graph for this problem is: 1 -> 5, 10 3 -> 4 -> 5 -> 3, 4 9 -> 5, 10 10 -> 9 Where: 1) f9#(I0, I1, I2) -> f2#(I0, I1, I2) 3) f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7] 4) f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0] 5) f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2] 9) f4#(I28, I29, I30) -> f2#(I28, I29, I30) 10) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] We have the following SCCs. { 9, 10 } DP problem for innermost termination. P = f4#(I28, I29, I30) -> f2#(I28, I29, I30) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] R = f10(x1, x2, x3) -> f1(x1, x2, x3) f9(I0, I1, I2) -> f2(I0, I1, I2) f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5] f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7] f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0] f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2] f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0] f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19] f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0] f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27] f4(I28, I29, I30) -> f2(I28, I29, I30) f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34] f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39] f1(I40, I41, I42) -> f2(I40, I41, I42)