/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I0, I1) f4#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5] f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 + I8 <= I9] f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= I12] f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0] f3#(I28, I29, I30, I31) -> f1#(I28, I29, I30, I31) [1 <= I30] f3#(I32, I33, I34, I35) -> f4#(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0] f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I0, I1) f4(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5] f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= I12] f5(I16, I17, I18, I19) -> f6(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f3(I20, I21, I22, I23) f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I26 <= 0] f3(I28, I29, I30, I31) -> f1(I28, I29, I30, I31) [1 <= I30] f3(I32, I33, I34, I35) -> f4(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0] f1(I36, I37, I38, I39) -> f2(I36, I37, -1 + I38, -1 + I39) The dependency graph for this problem is: 0 -> 1 1 -> 5 2 -> 3 -> 4 -> 5 -> 6, 7, 8 6 -> 9 7 -> 9 8 -> 2, 3, 4 9 -> 5 Where: 0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I0, I1) 2) f4#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5] 3) f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 + I8 <= I9] 4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= I12] 5) f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) 6) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0] 7) f3#(I28, I29, I30, I31) -> f1#(I28, I29, I30, I31) [1 <= I30] 8) f3#(I32, I33, I34, I35) -> f4#(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0] 9) f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) We have the following SCCs. { 5, 6, 7, 9 } DP problem for innermost termination. P = f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0] f3#(I28, I29, I30, I31) -> f1#(I28, I29, I30, I31) [1 <= I30] f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I0, I1) f4(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5] f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= I12] f5(I16, I17, I18, I19) -> f6(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f3(I20, I21, I22, I23) f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I26 <= 0] f3(I28, I29, I30, I31) -> f1(I28, I29, I30, I31) [1 <= I30] f3(I32, I33, I34, I35) -> f4(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0] f1(I36, I37, I38, I39) -> f2(I36, I37, -1 + I38, -1 + I39) We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3)] = x2 - 2 NU[f3#(x0,x1,x2,x3)] = x2 - 1 NU[f2#(x0,x1,x2,x3)] = x2 - 1 This gives the following inequalities: ==> I22 - 1 >= I22 - 1 1 + I26 <= 0 ==> I26 - 1 >= I26 - 2 1 <= I30 ==> I30 - 1 > I30 - 2 with I30 - 1 >= 0 ==> I38 - 2 >= (-1 + I38) - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0] f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f2(I0, I1, I0, I1) f4(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5] f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9] f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= I12] f5(I16, I17, I18, I19) -> f6(I16, I17, I18, I19) f2(I20, I21, I22, I23) -> f3(I20, I21, I22, I23) f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I26 <= 0] f3(I28, I29, I30, I31) -> f1(I28, I29, I30, I31) [1 <= I30] f3(I32, I33, I34, I35) -> f4(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0] f1(I36, I37, I38, I39) -> f2(I36, I37, -1 + I38, -1 + I39)