0.00/0.14 YES 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 cu#(true, x) -> cu#(x < 100000, x + 1) 0.00/0.14 R = 0.00/0.14 cu(true, x) -> cu(x < 100000, x + 1) 0.00/0.14 0.00/0.14 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 cu#(true, x) -> cu_1#(x) 0.00/0.14 cu_1#(x) -> cu#(x < 100000, x + 1) 0.00/0.14 cu_1#(x) -> cu_1#(x + 1) [x < 100000] 0.00/0.14 R = 0.00/0.14 cu(true, x) -> cu(x < 100000, x + 1) 0.00/0.14 0.00/0.14 The dependency graph for this problem is: 0.00/0.14 0 -> 2, 1 0.00/0.14 1 -> 0.00/0.14 2 -> 2, 1 0.00/0.14 Where: 0.00/0.14 0) cu#(true, x) -> cu_1#(x) 0.00/0.14 1) cu_1#(x) -> cu#(x < 100000, x + 1) 0.00/0.14 2) cu_1#(x) -> cu_1#(x + 1) [x < 100000] 0.00/0.14 0.00/0.14 We have the following SCCs. 0.00/0.14 { 2 } 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 cu_1#(x) -> cu_1#(x + 1) [x < 100000] 0.00/0.14 R = 0.00/0.14 cu(true, x) -> cu(x < 100000, x + 1) 0.00/0.14 0.00/0.14 We use the reverse value criterion with the projection function NU: 0.00/0.14 NU[cu_1#(z1)] = 100000 + -1 * z1 0.00/0.14 0.00/0.14 This gives the following inequalities: 0.00/0.14 x < 100000 ==> 100000 + -1 * x > 100000 + -1 * (x + 1) with 100000 + -1 * x >= 0 0.00/0.14 0.00/0.14 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.12 EOF