11.17/11.00 MAYBE 11.17/11.00 11.17/11.00 DP problem for innermost termination. 11.17/11.00 P = 11.17/11.00 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 11.17/11.00 f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] 11.17/11.00 f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1] 11.17/11.00 f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1] 11.17/11.00 f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] 11.17/11.00 f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1] 11.17/11.00 f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] 11.17/11.00 f1#(I18, I19, I20) -> f2#(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 11.17/11.00 R = 11.17/11.00 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 11.17/11.00 f3(I0, I1, I2) -> f2(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] 11.17/11.00 f2(I3, I4, I5) -> f2(0, 0, 9) [10 = I5 /\ I3 <= 1] 11.17/11.00 f2(I6, I7, I8) -> f2(1, 1, 2) [1 = I8 /\ I6 <= 1] 11.17/11.00 f3(I9, I10, I11) -> f2(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] 11.17/11.00 f2(I12, I13, I14) -> f3(I12, I14, I13) [10 <= I14 - 1] 11.17/11.00 f2(I15, I16, I17) -> f3(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] 11.17/11.00 f1(I18, I19, I20) -> f2(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 11.17/11.00 11.17/11.00 The dependency graph for this problem is: 11.17/11.00 0 -> 7 11.17/11.00 1 -> 2, 5, 6 11.17/11.00 2 -> 6 11.17/11.00 3 -> 6 11.17/11.00 4 -> 2, 3, 5, 6 11.17/11.00 5 -> 1, 4 11.17/11.00 6 -> 1, 4 11.17/11.00 7 -> 2, 5 11.17/11.00 Where: 11.17/11.00 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 11.17/11.00 1) f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] 11.17/11.00 2) f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1] 11.17/11.00 3) f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1] 11.17/11.00 4) f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] 11.17/11.00 5) f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1] 11.17/11.00 6) f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] 11.17/11.00 7) f1#(I18, I19, I20) -> f2#(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 11.17/11.00 11.17/11.00 We have the following SCCs. 11.17/11.00 { 1, 2, 3, 4, 5, 6 } 11.17/11.00 11.17/11.00 DP problem for innermost termination. 11.17/11.00 P = 11.17/11.00 f3#(I0, I1, I2) -> f2#(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] 11.17/11.00 f2#(I3, I4, I5) -> f2#(0, 0, 9) [10 = I5 /\ I3 <= 1] 11.17/11.00 f2#(I6, I7, I8) -> f2#(1, 1, 2) [1 = I8 /\ I6 <= 1] 11.17/11.00 f3#(I9, I10, I11) -> f2#(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] 11.17/11.00 f2#(I12, I13, I14) -> f3#(I12, I14, I13) [10 <= I14 - 1] 11.17/11.00 f2#(I15, I16, I17) -> f3#(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] 11.17/11.00 R = 11.17/11.00 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 11.17/11.00 f3(I0, I1, I2) -> f2(1, 1, I1 + 1) [1 = I2 /\ 1 <= I1 - 1 /\ I0 <= 1] 11.17/11.00 f2(I3, I4, I5) -> f2(0, 0, 9) [10 = I5 /\ I3 <= 1] 11.17/11.00 f2(I6, I7, I8) -> f2(1, 1, 2) [1 = I8 /\ I6 <= 1] 11.17/11.00 f3(I9, I10, I11) -> f2(0, 0, I10 - 1) [0 = I11 /\ 1 <= I10 - 1 /\ I9 <= 1] 11.17/11.00 f2(I12, I13, I14) -> f3(I12, I14, I13) [10 <= I14 - 1] 11.17/11.00 f2(I15, I16, I17) -> f3(I15, I17, I16) [1 <= I17 - 1 /\ I17 <= 9] 11.17/11.00 f1(I18, I19, I20) -> f2(0, 0, 10 * I19) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 11.17/11.00 11.17/13.98 EOF