0.97/1.04 MAYBE 0.97/1.04 0.97/1.04 DP problem for innermost termination. 0.97/1.04 P = 0.97/1.04 init#(x1, x2) -> f3#(rnd1, rnd2) 0.97/1.04 f4#(I0, I1) -> f4#(I0 - 1, I2) [0 <= I0 - 1] 0.97/1.04 f3#(I3, I4) -> f4#(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1] 0.97/1.04 f2#(I7, I8) -> f2#(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8] 0.97/1.04 f2#(I10, I11) -> f2#(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11] 0.97/1.04 f2#(I13, I14) -> f2#(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14] 0.97/1.04 f3#(I16, I17) -> f2#(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1] 0.97/1.04 f1#(I19, I20) -> f2#(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19] 0.97/1.04 R = 0.97/1.04 init(x1, x2) -> f3(rnd1, rnd2) 0.97/1.04 f4(I0, I1) -> f4(I0 - 1, I2) [0 <= I0 - 1] 0.97/1.04 f3(I3, I4) -> f4(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1] 0.97/1.04 f2(I7, I8) -> f2(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8] 0.97/1.04 f2(I10, I11) -> f2(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11] 0.97/1.04 f2(I13, I14) -> f2(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14] 0.97/1.04 f3(I16, I17) -> f2(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1] 0.97/1.04 f1(I19, I20) -> f2(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19] 0.97/1.04 0.97/1.04 The dependency graph for this problem is: 0.97/1.04 0 -> 2, 6 0.97/1.04 1 -> 1 0.97/1.04 2 -> 1 0.97/1.04 3 -> 5 0.97/1.04 4 -> 3 0.97/1.04 5 -> 4 0.97/1.04 6 -> 5 0.97/1.04 7 -> 5 0.97/1.04 Where: 0.97/1.04 0) init#(x1, x2) -> f3#(rnd1, rnd2) 0.97/1.04 1) f4#(I0, I1) -> f4#(I0 - 1, I2) [0 <= I0 - 1] 0.97/1.04 2) f3#(I3, I4) -> f4#(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1] 0.97/1.04 3) f2#(I7, I8) -> f2#(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8] 0.97/1.04 4) f2#(I10, I11) -> f2#(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11] 0.97/1.04 5) f2#(I13, I14) -> f2#(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14] 0.97/1.04 6) f3#(I16, I17) -> f2#(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1] 0.97/1.04 7) f1#(I19, I20) -> f2#(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19] 0.97/1.04 0.97/1.04 We have the following SCCs. 0.97/1.04 { 1 } 0.97/1.04 { 3, 4, 5 } 0.97/1.04 0.97/1.04 DP problem for innermost termination. 0.97/1.04 P = 0.97/1.04 f2#(I7, I8) -> f2#(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8] 0.97/1.04 f2#(I10, I11) -> f2#(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11] 0.97/1.04 f2#(I13, I14) -> f2#(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14] 0.97/1.04 R = 0.97/1.04 init(x1, x2) -> f3(rnd1, rnd2) 0.97/1.04 f4(I0, I1) -> f4(I0 - 1, I2) [0 <= I0 - 1] 0.97/1.04 f3(I3, I4) -> f4(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1] 0.97/1.04 f2(I7, I8) -> f2(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8] 0.97/1.04 f2(I10, I11) -> f2(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11] 0.97/1.04 f2(I13, I14) -> f2(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14] 0.97/1.04 f3(I16, I17) -> f2(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1] 0.97/1.04 f1(I19, I20) -> f2(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19] 0.97/1.04 0.97/4.02 EOF