7.03/7.01 MAYBE 7.03/7.01 7.03/7.01 DP problem for innermost termination. 7.03/7.01 P = 7.03/7.01 f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 7.03/7.01 f6#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 4, 0, 0, I6) 7.03/7.01 f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 7.03/7.01 f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 7.03/7.01 f3#(I29, I30, I31, I32, I33, I34, I35) -> f2#(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36] 7.03/7.01 f1#(I39, I40, I41, I42, I43, I44, I45) -> f2#(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46] 7.03/7.01 R = 7.03/7.01 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 7.03/7.01 f6(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 7.03/7.01 f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 7.03/7.01 f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 7.03/7.01 f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I28, I22, I26, I24, I25, I26, I27) [I28 = I26] 7.03/7.01 f3(I29, I30, I31, I32, I33, I34, I35) -> f2(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36] 7.03/7.01 f1(I39, I40, I41, I42, I43, I44, I45) -> f2(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46] 7.03/7.01 7.03/7.01 The dependency graph for this problem is: 7.03/7.01 0 -> 1 7.03/7.01 1 -> 3 7.03/7.01 2 -> 3 7.03/7.01 3 -> 2 7.03/7.01 4 -> 2 7.03/7.01 5 -> 2 7.03/7.01 Where: 7.03/7.01 0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 7.03/7.01 1) f6#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 4, 0, 0, I6) 7.03/7.01 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 7.03/7.01 3) f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 7.03/7.01 4) f3#(I29, I30, I31, I32, I33, I34, I35) -> f2#(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36] 7.03/7.01 5) f1#(I39, I40, I41, I42, I43, I44, I45) -> f2#(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46] 7.03/7.01 7.03/7.01 We have the following SCCs. 7.03/7.01 { 2, 3 } 7.03/7.01 7.03/7.01 DP problem for innermost termination. 7.03/7.01 P = 7.03/7.01 f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 7.03/7.01 f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 7.03/7.01 R = 7.03/7.01 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 7.03/7.01 f6(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 4, 0, 0, I6) 7.03/7.01 f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 7.03/7.01 f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1] 7.03/7.01 f4(I21, I22, I23, I24, I25, I26, I27) -> f5(I28, I22, I26, I24, I25, I26, I27) [I28 = I26] 7.03/7.01 f3(I29, I30, I31, I32, I33, I34, I35) -> f2(I36, I35, I31, I32, I33, I37, I35) [I38 = I35 /\ I37 = I38 /\ I36 = I36] 7.03/7.01 f1(I39, I40, I41, I42, I43, I44, I45) -> f2(I46, I45, I41, I42, I43, I47, I45) [I48 = I45 /\ I47 = I48 /\ I46 = I46] 7.03/7.01 7.03/9.99 EOF