6.30/6.26 MAYBE 6.30/6.26 6.30/6.26 DP problem for innermost termination. 6.30/6.26 P = 6.30/6.26 f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 6.30/6.26 f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] 6.30/6.26 f4#(I5, I6, I7, I8, I9) -> f3#(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] 6.30/6.26 f4#(I10, I11, I12, I13, I14) -> f2#(I10, I11, I12, I13, I14) [I14 <= 0] 6.30/6.26 f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, -1 + I19) [1 <= I15] 6.30/6.26 f3#(I20, I21, I22, I23, I24) -> f4#(I20, I21, I22, I23, -2 + I24) [I20 <= 0] 6.30/6.26 f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] 6.30/6.26 f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31] 6.30/6.26 f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0] 6.30/6.26 R = 6.30/6.26 f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 6.30/6.26 f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] 6.30/6.26 f4(I5, I6, I7, I8, I9) -> f3(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] 6.30/6.26 f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [I14 <= 0] 6.30/6.26 f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, -1 + I19) [1 <= I15] 6.30/6.26 f3(I20, I21, I22, I23, I24) -> f4(I20, I21, I22, I23, -2 + I24) [I20 <= 0] 6.30/6.26 f2(I25, I26, I27, I28, I29) -> f1(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] 6.30/6.26 f1(I30, I31, I32, I33, I34) -> f2(I30, I31, 1, I33, I34) [1 <= I31] 6.30/6.26 f1(I35, I36, I37, I38, I39) -> f2(I35, I36, 0, I38, I39) [I36 <= 0] 6.30/6.26 6.30/6.26 The dependency graph for this problem is: 6.30/6.26 0 -> 1 6.30/6.26 1 -> 2, 3 6.30/6.26 2 -> 4, 5 6.30/6.26 3 -> 6 6.30/6.26 4 -> 2, 3 6.30/6.26 5 -> 2, 3 6.30/6.26 6 -> 7, 8 6.30/6.26 7 -> 6 6.30/6.26 8 -> 6 6.30/6.26 Where: 6.30/6.26 0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 6.30/6.26 1) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] 6.30/6.26 2) f4#(I5, I6, I7, I8, I9) -> f3#(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] 6.30/6.26 3) f4#(I10, I11, I12, I13, I14) -> f2#(I10, I11, I12, I13, I14) [I14 <= 0] 6.30/6.26 4) f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, -1 + I19) [1 <= I15] 6.30/6.26 5) f3#(I20, I21, I22, I23, I24) -> f4#(I20, I21, I22, I23, -2 + I24) [I20 <= 0] 6.30/6.26 6) f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] 6.30/6.26 7) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31] 6.30/6.26 8) f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0] 6.30/6.26 6.30/6.26 We have the following SCCs. 6.30/6.26 { 2, 4, 5 } 6.30/6.26 { 6, 7, 8 } 6.30/6.26 6.30/6.26 DP problem for innermost termination. 6.30/6.26 P = 6.30/6.26 f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] 6.30/6.26 f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31] 6.30/6.26 f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0] 6.30/6.26 R = 6.30/6.26 f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 6.30/6.26 f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] 6.30/6.26 f4(I5, I6, I7, I8, I9) -> f3(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] 6.30/6.26 f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [I14 <= 0] 6.30/6.26 f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, -1 + I19) [1 <= I15] 6.30/6.26 f3(I20, I21, I22, I23, I24) -> f4(I20, I21, I22, I23, -2 + I24) [I20 <= 0] 6.30/6.26 f2(I25, I26, I27, I28, I29) -> f1(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] 6.30/6.26 f1(I30, I31, I32, I33, I34) -> f2(I30, I31, 1, I33, I34) [1 <= I31] 6.30/6.26 f1(I35, I36, I37, I38, I39) -> f2(I35, I36, 0, I38, I39) [I36 <= 0] 6.30/6.26 6.30/9.24 EOF