/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f19#(x1, x2, x3) -> f18#(x1, x2, x3) f18#(I0, I1, I2) -> f7#(I0, 0, I2) [y1 = 0] f3#(I3, I4, I5) -> f17#(I3, I4, I5) f3#(I6, I7, I8) -> f13#(I6, I7, I8) f3#(I9, I10, I11) -> f17#(I9, I10, I11) f17#(I12, I13, I14) -> f16#(I12, I13, I14) f16#(I15, I16, I17) -> f15#(I15, I16, I17) f16#(I18, I19, I20) -> f14#(I18, I19, I20) f16#(I21, I22, I23) -> f14#(I21, I22, I23) f15#(I24, I25, I26) -> f13#(I24, I25, I26) f14#(I27, I28, I29) -> f15#(I27, I28, I29) f2#(I30, I31, I32) -> f11#(I30, I31, I32) f13#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) f11#(I36, I37, I38) -> f10#(I36, I37, I38) [1 + I37 <= 100] f10#(I42, I43, I44) -> f9#(I42, I43, I44) f10#(I45, I46, I47) -> f4#(I45, I46, I47) f10#(I48, I49, I50) -> f9#(I48, I49, I50) f9#(I51, I52, I53) -> f8#(I51, I52, rnd3) [rnd3 = rnd3] f8#(I54, I55, I56) -> f6#(I54, I55, I56) f8#(I57, I58, I59) -> f5#(I57, I58, I59) f8#(I60, I61, I62) -> f5#(I60, I61, I62) f7#(I63, I64, I65) -> f1#(I63, I64, I65) f6#(I66, I67, I68) -> f4#(I66, I67, I68) f5#(I69, I70, I71) -> f6#(I69, I70, I71) f4#(I72, I73, I74) -> f2#(I72, 1 + I73, I74) f1#(I75, I76, I77) -> f3#(I76, I76, I77) [1 + I76 <= 100] f1#(I78, I79, I80) -> f2#(I78, 0, I80) [100 <= I79] R = f19(x1, x2, x3) -> f18(x1, x2, x3) f18(I0, I1, I2) -> f7(I0, 0, I2) [y1 = 0] f3(I3, I4, I5) -> f17(I3, I4, I5) f3(I6, I7, I8) -> f13(I6, I7, I8) f3(I9, I10, I11) -> f17(I9, I10, I11) f17(I12, I13, I14) -> f16(I12, I13, I14) f16(I15, I16, I17) -> f15(I15, I16, I17) f16(I18, I19, I20) -> f14(I18, I19, I20) f16(I21, I22, I23) -> f14(I21, I22, I23) f15(I24, I25, I26) -> f13(I24, I25, I26) f14(I27, I28, I29) -> f15(I27, I28, I29) f2(I30, I31, I32) -> f11(I30, I31, I32) f13(I33, I34, I35) -> f7(I33, 1 + I34, I35) f11(I36, I37, I38) -> f10(I36, I37, I38) [1 + I37 <= 100] f11(I39, I40, I41) -> f12(I39, I40, I41) [100 <= I40] f10(I42, I43, I44) -> f9(I42, I43, I44) f10(I45, I46, I47) -> f4(I45, I46, I47) f10(I48, I49, I50) -> f9(I48, I49, I50) f9(I51, I52, I53) -> f8(I51, I52, rnd3) [rnd3 = rnd3] f8(I54, I55, I56) -> f6(I54, I55, I56) f8(I57, I58, I59) -> f5(I57, I58, I59) f8(I60, I61, I62) -> f5(I60, I61, I62) f7(I63, I64, I65) -> f1(I63, I64, I65) f6(I66, I67, I68) -> f4(I66, I67, I68) f5(I69, I70, I71) -> f6(I69, I70, I71) f4(I72, I73, I74) -> f2(I72, 1 + I73, I74) f1(I75, I76, I77) -> f3(I76, I76, I77) [1 + I76 <= 100] f1(I78, I79, I80) -> f2(I78, 0, I80) [100 <= I79] The dependency graph for this problem is: 0 -> 1 1 -> 21 2 -> 5 3 -> 12 4 -> 5 5 -> 6, 7, 8 6 -> 9 7 -> 10 8 -> 10 9 -> 12 10 -> 9 11 -> 13 12 -> 21 13 -> 14, 15, 16 14 -> 17 15 -> 24 16 -> 17 17 -> 18, 19, 20 18 -> 22 19 -> 23 20 -> 23 21 -> 25, 26 22 -> 24 23 -> 22 24 -> 11 25 -> 2, 3, 4 26 -> 11 Where: 0) f19#(x1, x2, x3) -> f18#(x1, x2, x3) 1) f18#(I0, I1, I2) -> f7#(I0, 0, I2) [y1 = 0] 2) f3#(I3, I4, I5) -> f17#(I3, I4, I5) 3) f3#(I6, I7, I8) -> f13#(I6, I7, I8) 4) f3#(I9, I10, I11) -> f17#(I9, I10, I11) 5) f17#(I12, I13, I14) -> f16#(I12, I13, I14) 6) f16#(I15, I16, I17) -> f15#(I15, I16, I17) 7) f16#(I18, I19, I20) -> f14#(I18, I19, I20) 8) f16#(I21, I22, I23) -> f14#(I21, I22, I23) 9) f15#(I24, I25, I26) -> f13#(I24, I25, I26) 10) f14#(I27, I28, I29) -> f15#(I27, I28, I29) 11) f2#(I30, I31, I32) -> f11#(I30, I31, I32) 12) f13#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) 13) f11#(I36, I37, I38) -> f10#(I36, I37, I38) [1 + I37 <= 100] 14) f10#(I42, I43, I44) -> f9#(I42, I43, I44) 15) f10#(I45, I46, I47) -> f4#(I45, I46, I47) 16) f10#(I48, I49, I50) -> f9#(I48, I49, I50) 17) f9#(I51, I52, I53) -> f8#(I51, I52, rnd3) [rnd3 = rnd3] 18) f8#(I54, I55, I56) -> f6#(I54, I55, I56) 19) f8#(I57, I58, I59) -> f5#(I57, I58, I59) 20) f8#(I60, I61, I62) -> f5#(I60, I61, I62) 21) f7#(I63, I64, I65) -> f1#(I63, I64, I65) 22) f6#(I66, I67, I68) -> f4#(I66, I67, I68) 23) f5#(I69, I70, I71) -> f6#(I69, I70, I71) 24) f4#(I72, I73, I74) -> f2#(I72, 1 + I73, I74) 25) f1#(I75, I76, I77) -> f3#(I76, I76, I77) [1 + I76 <= 100] 26) f1#(I78, I79, I80) -> f2#(I78, 0, I80) [100 <= I79] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 21, 25 } { 11, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24 } DP problem for innermost termination. P = f2#(I30, I31, I32) -> f11#(I30, I31, I32) f11#(I36, I37, I38) -> f10#(I36, I37, I38) [1 + I37 <= 100] f10#(I42, I43, I44) -> f9#(I42, I43, I44) f10#(I45, I46, I47) -> f4#(I45, I46, I47) f10#(I48, I49, I50) -> f9#(I48, I49, I50) f9#(I51, I52, I53) -> f8#(I51, I52, rnd3) [rnd3 = rnd3] f8#(I54, I55, I56) -> f6#(I54, I55, I56) f8#(I57, I58, I59) -> f5#(I57, I58, I59) f8#(I60, I61, I62) -> f5#(I60, I61, I62) f6#(I66, I67, I68) -> f4#(I66, I67, I68) f5#(I69, I70, I71) -> f6#(I69, I70, I71) f4#(I72, I73, I74) -> f2#(I72, 1 + I73, I74) R = f19(x1, x2, x3) -> f18(x1, x2, x3) f18(I0, I1, I2) -> f7(I0, 0, I2) [y1 = 0] f3(I3, I4, I5) -> f17(I3, I4, I5) f3(I6, I7, I8) -> f13(I6, I7, I8) f3(I9, I10, I11) -> f17(I9, I10, I11) f17(I12, I13, I14) -> f16(I12, I13, I14) f16(I15, I16, I17) -> f15(I15, I16, I17) f16(I18, I19, I20) -> f14(I18, I19, I20) f16(I21, I22, I23) -> f14(I21, I22, I23) f15(I24, I25, I26) -> f13(I24, I25, I26) f14(I27, I28, I29) -> f15(I27, I28, I29) f2(I30, I31, I32) -> f11(I30, I31, I32) f13(I33, I34, I35) -> f7(I33, 1 + I34, I35) f11(I36, I37, I38) -> f10(I36, I37, I38) [1 + I37 <= 100] f11(I39, I40, I41) -> f12(I39, I40, I41) [100 <= I40] f10(I42, I43, I44) -> f9(I42, I43, I44) f10(I45, I46, I47) -> f4(I45, I46, I47) f10(I48, I49, I50) -> f9(I48, I49, I50) f9(I51, I52, I53) -> f8(I51, I52, rnd3) [rnd3 = rnd3] f8(I54, I55, I56) -> f6(I54, I55, I56) f8(I57, I58, I59) -> f5(I57, I58, I59) f8(I60, I61, I62) -> f5(I60, I61, I62) f7(I63, I64, I65) -> f1(I63, I64, I65) f6(I66, I67, I68) -> f4(I66, I67, I68) f5(I69, I70, I71) -> f6(I69, I70, I71) f4(I72, I73, I74) -> f2(I72, 1 + I73, I74) f1(I75, I76, I77) -> f3(I76, I76, I77) [1 + I76 <= 100] f1(I78, I79, I80) -> f2(I78, 0, I80) [100 <= I79] We use the extended value criterion with the projection function NU: NU[f5#(x0,x1,x2)] = -x1 + 98 NU[f6#(x0,x1,x2)] = -x1 + 98 NU[f8#(x0,x1,x2)] = -x1 + 98 NU[f4#(x0,x1,x2)] = -x1 + 98 NU[f9#(x0,x1,x2)] = -x1 + 98 NU[f10#(x0,x1,x2)] = -x1 + 98 NU[f11#(x0,x1,x2)] = -x1 + 99 NU[f2#(x0,x1,x2)] = -x1 + 99 This gives the following inequalities: ==> -I31 + 99 >= -I31 + 99 1 + I37 <= 100 ==> -I37 + 99 > -I37 + 98 with -I37 + 99 >= 0 ==> -I43 + 98 >= -I43 + 98 ==> -I46 + 98 >= -I46 + 98 ==> -I49 + 98 >= -I49 + 98 rnd3 = rnd3 ==> -I52 + 98 >= -I52 + 98 ==> -I55 + 98 >= -I55 + 98 ==> -I58 + 98 >= -I58 + 98 ==> -I61 + 98 >= -I61 + 98 ==> -I67 + 98 >= -I67 + 98 ==> -I70 + 98 >= -I70 + 98 ==> -I73 + 98 >= -(1 + I73) + 99 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I30, I31, I32) -> f11#(I30, I31, I32) f10#(I42, I43, I44) -> f9#(I42, I43, I44) f10#(I45, I46, I47) -> f4#(I45, I46, I47) f10#(I48, I49, I50) -> f9#(I48, I49, I50) f9#(I51, I52, I53) -> f8#(I51, I52, rnd3) [rnd3 = rnd3] f8#(I54, I55, I56) -> f6#(I54, I55, I56) f8#(I57, I58, I59) -> f5#(I57, I58, I59) f8#(I60, I61, I62) -> f5#(I60, I61, I62) f6#(I66, I67, I68) -> f4#(I66, I67, I68) f5#(I69, I70, I71) -> f6#(I69, I70, I71) f4#(I72, I73, I74) -> f2#(I72, 1 + I73, I74) R = f19(x1, x2, x3) -> f18(x1, x2, x3) f18(I0, I1, I2) -> f7(I0, 0, I2) [y1 = 0] f3(I3, I4, I5) -> f17(I3, I4, I5) f3(I6, I7, I8) -> f13(I6, I7, I8) f3(I9, I10, I11) -> f17(I9, I10, I11) f17(I12, I13, I14) -> f16(I12, I13, I14) f16(I15, I16, I17) -> f15(I15, I16, I17) f16(I18, I19, I20) -> f14(I18, I19, I20) f16(I21, I22, I23) -> f14(I21, I22, I23) f15(I24, I25, I26) -> f13(I24, I25, I26) f14(I27, I28, I29) -> f15(I27, I28, I29) f2(I30, I31, I32) -> f11(I30, I31, I32) f13(I33, I34, I35) -> f7(I33, 1 + I34, I35) f11(I36, I37, I38) -> f10(I36, I37, I38) [1 + I37 <= 100] f11(I39, I40, I41) -> f12(I39, I40, I41) [100 <= I40] f10(I42, I43, I44) -> f9(I42, I43, I44) f10(I45, I46, I47) -> f4(I45, I46, I47) f10(I48, I49, I50) -> f9(I48, I49, I50) f9(I51, I52, I53) -> f8(I51, I52, rnd3) [rnd3 = rnd3] f8(I54, I55, I56) -> f6(I54, I55, I56) f8(I57, I58, I59) -> f5(I57, I58, I59) f8(I60, I61, I62) -> f5(I60, I61, I62) f7(I63, I64, I65) -> f1(I63, I64, I65) f6(I66, I67, I68) -> f4(I66, I67, I68) f5(I69, I70, I71) -> f6(I69, I70, I71) f4(I72, I73, I74) -> f2(I72, 1 + I73, I74) f1(I75, I76, I77) -> f3(I76, I76, I77) [1 + I76 <= 100] f1(I78, I79, I80) -> f2(I78, 0, I80) [100 <= I79] The dependency graph for this problem is: 11 -> 14 -> 17 15 -> 24 16 -> 17 17 -> 18, 19, 20 18 -> 22 19 -> 23 20 -> 23 22 -> 24 23 -> 22 24 -> 11 Where: 11) f2#(I30, I31, I32) -> f11#(I30, I31, I32) 14) f10#(I42, I43, I44) -> f9#(I42, I43, I44) 15) f10#(I45, I46, I47) -> f4#(I45, I46, I47) 16) f10#(I48, I49, I50) -> f9#(I48, I49, I50) 17) f9#(I51, I52, I53) -> f8#(I51, I52, rnd3) [rnd3 = rnd3] 18) f8#(I54, I55, I56) -> f6#(I54, I55, I56) 19) f8#(I57, I58, I59) -> f5#(I57, I58, I59) 20) f8#(I60, I61, I62) -> f5#(I60, I61, I62) 22) f6#(I66, I67, I68) -> f4#(I66, I67, I68) 23) f5#(I69, I70, I71) -> f6#(I69, I70, I71) 24) f4#(I72, I73, I74) -> f2#(I72, 1 + I73, I74) We have the following SCCs. DP problem for innermost termination. P = f3#(I3, I4, I5) -> f17#(I3, I4, I5) f3#(I6, I7, I8) -> f13#(I6, I7, I8) f3#(I9, I10, I11) -> f17#(I9, I10, I11) f17#(I12, I13, I14) -> f16#(I12, I13, I14) f16#(I15, I16, I17) -> f15#(I15, I16, I17) f16#(I18, I19, I20) -> f14#(I18, I19, I20) f16#(I21, I22, I23) -> f14#(I21, I22, I23) f15#(I24, I25, I26) -> f13#(I24, I25, I26) f14#(I27, I28, I29) -> f15#(I27, I28, I29) f13#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) f7#(I63, I64, I65) -> f1#(I63, I64, I65) f1#(I75, I76, I77) -> f3#(I76, I76, I77) [1 + I76 <= 100] R = f19(x1, x2, x3) -> f18(x1, x2, x3) f18(I0, I1, I2) -> f7(I0, 0, I2) [y1 = 0] f3(I3, I4, I5) -> f17(I3, I4, I5) f3(I6, I7, I8) -> f13(I6, I7, I8) f3(I9, I10, I11) -> f17(I9, I10, I11) f17(I12, I13, I14) -> f16(I12, I13, I14) f16(I15, I16, I17) -> f15(I15, I16, I17) f16(I18, I19, I20) -> f14(I18, I19, I20) f16(I21, I22, I23) -> f14(I21, I22, I23) f15(I24, I25, I26) -> f13(I24, I25, I26) f14(I27, I28, I29) -> f15(I27, I28, I29) f2(I30, I31, I32) -> f11(I30, I31, I32) f13(I33, I34, I35) -> f7(I33, 1 + I34, I35) f11(I36, I37, I38) -> f10(I36, I37, I38) [1 + I37 <= 100] f11(I39, I40, I41) -> f12(I39, I40, I41) [100 <= I40] f10(I42, I43, I44) -> f9(I42, I43, I44) f10(I45, I46, I47) -> f4(I45, I46, I47) f10(I48, I49, I50) -> f9(I48, I49, I50) f9(I51, I52, I53) -> f8(I51, I52, rnd3) [rnd3 = rnd3] f8(I54, I55, I56) -> f6(I54, I55, I56) f8(I57, I58, I59) -> f5(I57, I58, I59) f8(I60, I61, I62) -> f5(I60, I61, I62) f7(I63, I64, I65) -> f1(I63, I64, I65) f6(I66, I67, I68) -> f4(I66, I67, I68) f5(I69, I70, I71) -> f6(I69, I70, I71) f4(I72, I73, I74) -> f2(I72, 1 + I73, I74) f1(I75, I76, I77) -> f3(I76, I76, I77) [1 + I76 <= 100] f1(I78, I79, I80) -> f2(I78, 0, I80) [100 <= I79] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2)] = -x1 + 99 NU[f7#(x0,x1,x2)] = -x1 + 99 NU[f14#(x0,x1,x2)] = -x1 + 98 NU[f15#(x0,x1,x2)] = -x1 + 98 NU[f16#(x0,x1,x2)] = -x1 + 98 NU[f13#(x0,x1,x2)] = -x1 + 98 NU[f17#(x0,x1,x2)] = -x1 + 98 NU[f3#(x0,x1,x2)] = -x1 + 98 This gives the following inequalities: ==> -I4 + 98 >= -I4 + 98 ==> -I7 + 98 >= -I7 + 98 ==> -I10 + 98 >= -I10 + 98 ==> -I13 + 98 >= -I13 + 98 ==> -I16 + 98 >= -I16 + 98 ==> -I19 + 98 >= -I19 + 98 ==> -I22 + 98 >= -I22 + 98 ==> -I25 + 98 >= -I25 + 98 ==> -I28 + 98 >= -I28 + 98 ==> -I34 + 98 >= -(1 + I34) + 99 ==> -I64 + 99 >= -I64 + 99 1 + I76 <= 100 ==> -I76 + 99 > -I76 + 98 with -I76 + 99 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I3, I4, I5) -> f17#(I3, I4, I5) f3#(I6, I7, I8) -> f13#(I6, I7, I8) f3#(I9, I10, I11) -> f17#(I9, I10, I11) f17#(I12, I13, I14) -> f16#(I12, I13, I14) f16#(I15, I16, I17) -> f15#(I15, I16, I17) f16#(I18, I19, I20) -> f14#(I18, I19, I20) f16#(I21, I22, I23) -> f14#(I21, I22, I23) f15#(I24, I25, I26) -> f13#(I24, I25, I26) f14#(I27, I28, I29) -> f15#(I27, I28, I29) f13#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) f7#(I63, I64, I65) -> f1#(I63, I64, I65) R = f19(x1, x2, x3) -> f18(x1, x2, x3) f18(I0, I1, I2) -> f7(I0, 0, I2) [y1 = 0] f3(I3, I4, I5) -> f17(I3, I4, I5) f3(I6, I7, I8) -> f13(I6, I7, I8) f3(I9, I10, I11) -> f17(I9, I10, I11) f17(I12, I13, I14) -> f16(I12, I13, I14) f16(I15, I16, I17) -> f15(I15, I16, I17) f16(I18, I19, I20) -> f14(I18, I19, I20) f16(I21, I22, I23) -> f14(I21, I22, I23) f15(I24, I25, I26) -> f13(I24, I25, I26) f14(I27, I28, I29) -> f15(I27, I28, I29) f2(I30, I31, I32) -> f11(I30, I31, I32) f13(I33, I34, I35) -> f7(I33, 1 + I34, I35) f11(I36, I37, I38) -> f10(I36, I37, I38) [1 + I37 <= 100] f11(I39, I40, I41) -> f12(I39, I40, I41) [100 <= I40] f10(I42, I43, I44) -> f9(I42, I43, I44) f10(I45, I46, I47) -> f4(I45, I46, I47) f10(I48, I49, I50) -> f9(I48, I49, I50) f9(I51, I52, I53) -> f8(I51, I52, rnd3) [rnd3 = rnd3] f8(I54, I55, I56) -> f6(I54, I55, I56) f8(I57, I58, I59) -> f5(I57, I58, I59) f8(I60, I61, I62) -> f5(I60, I61, I62) f7(I63, I64, I65) -> f1(I63, I64, I65) f6(I66, I67, I68) -> f4(I66, I67, I68) f5(I69, I70, I71) -> f6(I69, I70, I71) f4(I72, I73, I74) -> f2(I72, 1 + I73, I74) f1(I75, I76, I77) -> f3(I76, I76, I77) [1 + I76 <= 100] f1(I78, I79, I80) -> f2(I78, 0, I80) [100 <= I79] The dependency graph for this problem is: 2 -> 5 3 -> 12 4 -> 5 5 -> 6, 7, 8 6 -> 9 7 -> 10 8 -> 10 9 -> 12 10 -> 9 12 -> 21 21 -> Where: 2) f3#(I3, I4, I5) -> f17#(I3, I4, I5) 3) f3#(I6, I7, I8) -> f13#(I6, I7, I8) 4) f3#(I9, I10, I11) -> f17#(I9, I10, I11) 5) f17#(I12, I13, I14) -> f16#(I12, I13, I14) 6) f16#(I15, I16, I17) -> f15#(I15, I16, I17) 7) f16#(I18, I19, I20) -> f14#(I18, I19, I20) 8) f16#(I21, I22, I23) -> f14#(I21, I22, I23) 9) f15#(I24, I25, I26) -> f13#(I24, I25, I26) 10) f14#(I27, I28, I29) -> f15#(I27, I28, I29) 12) f13#(I33, I34, I35) -> f7#(I33, 1 + I34, I35) 21) f7#(I63, I64, I65) -> f1#(I63, I64, I65) We have the following SCCs.