9.50/9.61 YES 9.50/9.61 9.50/9.61 DP problem for innermost termination. 9.50/9.61 P = 9.50/9.61 f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 9.50/9.61 f8#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 4, 0, 0, I6) 9.50/9.61 f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 9.50/9.61 f4#(I37, I38, I39, I40, I41, I42, I43) -> f5#(I37, I38, I39, I40, I41, I42, I43) [1 <= I42] 9.50/9.61 f4#(I44, I45, I46, I47, I48, I49, I50) -> f5#(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0] 9.50/9.61 f3#(I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58] 9.50/9.61 f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68] 9.50/9.61 R = 9.50/9.61 f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 9.50/9.61 f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 9.50/9.61 f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 9.50/9.61 f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0] 9.50/9.61 f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34] 9.50/9.61 f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42] 9.50/9.61 f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0] 9.50/9.61 f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58] 9.50/9.61 f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68] 9.50/9.61 9.50/9.61 The dependency graph for this problem is: 9.50/9.61 0 -> 1 9.50/9.61 1 -> 3 9.50/9.61 2 -> 3, 4, 5 9.50/9.61 3 -> 2 9.50/9.61 4 -> 9.50/9.61 5 -> 9.50/9.61 6 -> 3, 4, 5 9.50/9.61 7 -> 2 9.50/9.61 Where: 9.50/9.61 0) f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 9.50/9.61 1) f8#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 4, 0, 0, I6) 9.50/9.61 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 3) f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 9.50/9.61 4) f4#(I37, I38, I39, I40, I41, I42, I43) -> f5#(I37, I38, I39, I40, I41, I42, I43) [1 <= I42] 9.50/9.61 5) f4#(I44, I45, I46, I47, I48, I49, I50) -> f5#(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0] 9.50/9.61 6) f3#(I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58] 9.50/9.61 7) f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68] 9.50/9.61 9.50/9.61 We have the following SCCs. 9.50/9.61 { 2, 3 } 9.50/9.61 9.50/9.61 DP problem for innermost termination. 9.50/9.61 P = 9.50/9.61 f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 9.50/9.61 R = 9.50/9.61 f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 9.50/9.61 f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 9.50/9.61 f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 9.50/9.61 f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0] 9.50/9.61 f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34] 9.50/9.61 f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42] 9.50/9.61 f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0] 9.50/9.61 f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58] 9.50/9.61 f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68] 9.50/9.61 9.50/9.61 We use the reverse value criterion with the projection function NU: 9.50/9.61 NU[f4#(z1,z2,z3,z4,z5,z6,z7)] = -1 + z4 - z5 + -1 * 0 9.50/9.61 NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = -1 + z4 - (1 + z5) + -1 * 0 9.50/9.61 9.50/9.61 This gives the following inequalities: 9.50/9.61 ==> -1 + I10 - (1 + I11) + -1 * 0 >= -1 + I10 - (1 + I11) + -1 * 0 9.50/9.61 I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1 ==> -1 + I17 - I18 + -1 * 0 > -1 + I17 - (1 + I18) + -1 * 0 with -1 + I17 - I18 + -1 * 0 >= 0 9.50/9.61 9.50/9.61 We remove all the strictly oriented dependency pairs. 9.50/9.61 9.50/9.61 DP problem for innermost termination. 9.50/9.61 P = 9.50/9.61 f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 R = 9.50/9.61 f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 9.50/9.61 f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 9.50/9.61 f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 9.50/9.61 f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0] 9.50/9.61 f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34] 9.50/9.61 f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42] 9.50/9.61 f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0] 9.50/9.61 f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58] 9.50/9.61 f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68] 9.50/9.61 9.50/9.61 The dependency graph for this problem is: 9.50/9.61 2 -> 9.50/9.61 Where: 9.50/9.61 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 9.50/9.61 9.50/9.61 We have the following SCCs. 9.50/9.61 9.50/12.58 EOF