0.00/0.11 YES 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 if2#(false, x, y) -> f#(x, y) 0.00/0.11 if1#(true, I4, I5) -> h#(I4, I5) 0.00/0.11 h#(I6, I7) -> if2#(I6 > I7, I6, I7) 0.00/0.11 f#(I8, I9) -> if1#(I8 > I9, I8, I9) 0.00/0.11 R = 0.00/0.11 if2(false, x, y) -> f(x, y) 0.00/0.11 if2(true, I0, I1) -> 0 0.00/0.11 if1(false, I2, I3) -> 0 0.00/0.11 if1(true, I4, I5) -> h(I4, I5) 0.00/0.11 h(I6, I7) -> if2(I6 > I7, I6, I7) 0.00/0.11 f(I8, I9) -> if1(I8 > I9, I8, I9) 0.00/0.11 0.00/0.11 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.11 0.00/0.11 DP problem for innermost termination. 0.00/0.11 P = 0.00/0.11 if2#(false, x, y) -> f#(x, y) 0.00/0.11 if1#(true, I4, I5) -> h#(I4, I5) 0.00/0.11 h#(I6, I7) -> if2#(I6 > I7, I6, I7) 0.00/0.11 f#(I8, I9) -> if1#(I8 > I9, I8, I9) 0.00/0.11 f#(I8, I9) -> h#(I8, I9) [I8 > I9] 0.00/0.11 h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)] 0.00/0.11 R = 0.00/0.11 if2(false, x, y) -> f(x, y) 0.00/0.11 if2(true, I0, I1) -> 0 0.00/0.11 if1(false, I2, I3) -> 0 0.00/0.11 if1(true, I4, I5) -> h(I4, I5) 0.00/0.11 h(I6, I7) -> if2(I6 > I7, I6, I7) 0.00/0.11 f(I8, I9) -> if1(I8 > I9, I8, I9) 0.00/0.11 0.00/0.11 The dependency graph for this problem is: 0.00/0.11 0 -> 4, 3 0.00/0.11 1 -> 5, 2 0.00/0.11 2 -> 0.00/0.11 3 -> 0.00/0.11 4 -> 2 0.00/0.11 5 -> 3 0.00/0.11 Where: 0.00/0.11 0) if2#(false, x, y) -> f#(x, y) 0.00/0.11 1) if1#(true, I4, I5) -> h#(I4, I5) 0.00/0.11 2) h#(I6, I7) -> if2#(I6 > I7, I6, I7) 0.00/0.11 3) f#(I8, I9) -> if1#(I8 > I9, I8, I9) 0.00/0.11 4) f#(I8, I9) -> h#(I8, I9) [I8 > I9] 0.00/0.11 5) h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)] 0.00/0.11 0.00/0.11 We have the following SCCs. 0.00/0.11 0.00/3.09 EOF