0.00/0.25 MAYBE 0.00/0.25 0.00/0.25 DP problem for innermost termination. 0.00/0.25 P = 0.00/0.25 min#(I2, I3) -> if#(I2 < I3, I2, I3) 0.00/0.25 cond#(y, x, y) -> minus#(x, y + 1) 0.00/0.25 minus#(I4, I5) -> cond#(min(I4, I5), I4, I5) 0.00/0.25 minus#(I4, I5) -> min#(I4, I5) 0.00/0.25 R = 0.00/0.25 if(false, u, v) -> v 0.00/0.25 if(true, I0, I1) -> I0 0.00/0.25 min(I2, I3) -> if(I2 < I3, I2, I3) 0.00/0.25 cond(y, x, y) -> 1 + minus(x, y + 1) 0.00/0.25 minus(I4, I5) -> cond(min(I4, I5), I4, I5) 0.00/0.25 minus(I6, I6) -> 0 0.00/0.25 0.00/0.25 The dependency graph for this problem is: 0.00/0.25 0 -> 0.00/0.25 1 -> 2, 3 0.00/0.25 2 -> 1 0.00/0.25 3 -> 0 0.00/0.25 Where: 0.00/0.25 0) min#(I2, I3) -> if#(I2 < I3, I2, I3) 0.00/0.25 1) cond#(y, x, y) -> minus#(x, y + 1) 0.00/0.25 2) minus#(I4, I5) -> cond#(min(I4, I5), I4, I5) 0.00/0.25 3) minus#(I4, I5) -> min#(I4, I5) 0.00/0.25 0.00/0.25 We have the following SCCs. 0.00/0.25 { 1, 2 } 0.00/0.25 0.00/0.25 DP problem for innermost termination. 0.00/0.25 P = 0.00/0.25 cond#(y, x, y) -> minus#(x, y + 1) 0.00/0.25 minus#(I4, I5) -> cond#(min(I4, I5), I4, I5) 0.00/0.25 R = 0.00/0.25 if(false, u, v) -> v 0.00/0.25 if(true, I0, I1) -> I0 0.00/0.25 min(I2, I3) -> if(I2 < I3, I2, I3) 0.00/0.25 cond(y, x, y) -> 1 + minus(x, y + 1) 0.00/0.25 minus(I4, I5) -> cond(min(I4, I5), I4, I5) 0.00/0.25 minus(I6, I6) -> 0 0.00/0.25 0.00/3.23 EOF