0.00/0.33 MAYBE 0.00/0.33 0.00/0.33 DP problem for innermost termination. 0.00/0.33 P = 0.00/0.33 condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 0.00/0.33 sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 0.00/0.33 sqrt#(I4) -> sqrtAcc#(I4, 0) 0.00/0.33 R = 0.00/0.33 condAcc(false, x, y) -> sqrtAcc(x, y + 1) 0.00/0.33 condAcc(true, I0, I1) -> I1 0.00/0.33 sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) 0.00/0.33 sqrt(I4) -> sqrtAcc(I4, 0) 0.00/0.33 0.00/0.33 This problem is converted using chaining, where edges between chained DPs are removed. 0.00/0.33 0.00/0.33 DP problem for innermost termination. 0.00/0.33 P = 0.00/0.33 condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 0.00/0.33 sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 0.00/0.33 sqrt#(I4) -> sqrtAcc#(I4, 0) 0.00/0.33 sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)] 0.00/0.33 R = 0.00/0.33 condAcc(false, x, y) -> sqrtAcc(x, y + 1) 0.00/0.33 condAcc(true, I0, I1) -> I1 0.00/0.33 sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) 0.00/0.33 sqrt(I4) -> sqrtAcc(I4, 0) 0.00/0.33 0.00/0.33 The dependency graph for this problem is: 0.00/0.33 0 -> 3, 1 0.00/0.33 1 -> 0.00/0.33 2 -> 3, 1 0.00/0.33 3 -> 3, 1 0.00/0.33 Where: 0.00/0.33 0) condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 0.00/0.33 1) sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 0.00/0.33 2) sqrt#(I4) -> sqrtAcc#(I4, 0) 0.00/0.33 3) sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)] 0.00/0.33 0.00/0.33 We have the following SCCs. 0.00/0.33 { 3 } 0.00/0.33 0.00/0.33 DP problem for innermost termination. 0.00/0.33 P = 0.00/0.33 sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)] 0.00/0.33 R = 0.00/0.33 condAcc(false, x, y) -> sqrtAcc(x, y + 1) 0.00/0.33 condAcc(true, I0, I1) -> I1 0.00/0.33 sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) 0.00/0.33 sqrt(I4) -> sqrtAcc(I4, 0) 0.00/0.33 0.00/3.31 EOF