5.34/5.29	MAYBE
5.34/5.29	
5.34/5.29	DP problem for innermost termination.
5.34/5.29	P =
5.34/5.29	  f7#(x1, x2) -> f6#(x1, x2) 
5.34/5.29	  f6#(I0, I1) -> f2#(0, I1) 
5.34/5.29	  f2#(I2, I3) -> f4#(I2, rnd2) [rnd2 = rnd2]
5.34/5.29	  f4#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 0]
5.34/5.29	  f4#(I6, I7) -> f3#(I6, I7) [1 <= I7]
5.34/5.29	  f3#(I10, I11) -> f1#(0, I11) [4 <= I10 /\ I10 <= 4]
5.34/5.29	  f3#(I12, I13) -> f1#(I12, I13) [1 + I12 <= 4]
5.34/5.29	  f3#(I14, I15) -> f1#(I14, I15) [5 <= I14]
5.34/5.29	  f1#(I16, I17) -> f2#(1 + I16, I17) 
5.34/5.29	R = 
5.34/5.29	  f7(x1, x2) -> f6(x1, x2) 
5.34/5.29	  f6(I0, I1) -> f2(0, I1) 
5.34/5.29	  f2(I2, I3) -> f4(I2, rnd2) [rnd2 = rnd2]
5.34/5.29	  f4(I4, I5) -> f3(I4, I5) [1 + I5 <= 0]
5.34/5.29	  f4(I6, I7) -> f3(I6, I7) [1 <= I7]
5.34/5.29	  f4(I8, I9) -> f5(I8, I9) [0 <= I9 /\ I9 <= 0]
5.34/5.29	  f3(I10, I11) -> f1(0, I11) [4 <= I10 /\ I10 <= 4]
5.34/5.29	  f3(I12, I13) -> f1(I12, I13) [1 + I12 <= 4]
5.34/5.29	  f3(I14, I15) -> f1(I14, I15) [5 <= I14]
5.34/5.29	  f1(I16, I17) -> f2(1 + I16, I17) 
5.34/5.29	
5.34/5.29	The dependency graph for this problem is:
5.34/5.29	  0 -> 1
5.34/5.29	  1 -> 2
5.34/5.29	  2 -> 3, 4
5.34/5.29	  3 -> 5, 6, 7
5.34/5.29	  4 -> 5, 6, 7
5.34/5.29	  5 -> 8
5.34/5.29	  6 -> 8
5.34/5.29	  7 -> 8
5.34/5.29	  8 -> 2
5.34/5.29	Where:
5.34/5.29	  0) f7#(x1, x2) -> f6#(x1, x2) 
5.34/5.29	  1) f6#(I0, I1) -> f2#(0, I1) 
5.34/5.30	  2) f2#(I2, I3) -> f4#(I2, rnd2) [rnd2 = rnd2]
5.34/5.30	  3) f4#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 0]
5.34/5.30	  4) f4#(I6, I7) -> f3#(I6, I7) [1 <= I7]
5.34/5.30	  5) f3#(I10, I11) -> f1#(0, I11) [4 <= I10 /\ I10 <= 4]
5.34/5.30	  6) f3#(I12, I13) -> f1#(I12, I13) [1 + I12 <= 4]
5.34/5.30	  7) f3#(I14, I15) -> f1#(I14, I15) [5 <= I14]
5.34/5.30	  8) f1#(I16, I17) -> f2#(1 + I16, I17) 
5.34/5.30	
5.34/5.30	We have the following SCCs.
5.34/5.30	  { 2, 3, 4, 5, 6, 7, 8 }
5.34/5.30	
5.34/5.30	DP problem for innermost termination.
5.34/5.30	P =
5.34/5.30	  f2#(I2, I3) -> f4#(I2, rnd2) [rnd2 = rnd2]
5.34/5.30	  f4#(I4, I5) -> f3#(I4, I5) [1 + I5 <= 0]
5.34/5.30	  f4#(I6, I7) -> f3#(I6, I7) [1 <= I7]
5.34/5.30	  f3#(I10, I11) -> f1#(0, I11) [4 <= I10 /\ I10 <= 4]
5.34/5.30	  f3#(I12, I13) -> f1#(I12, I13) [1 + I12 <= 4]
5.34/5.30	  f3#(I14, I15) -> f1#(I14, I15) [5 <= I14]
5.34/5.30	  f1#(I16, I17) -> f2#(1 + I16, I17) 
5.34/5.30	R = 
5.34/5.30	  f7(x1, x2) -> f6(x1, x2) 
5.34/5.30	  f6(I0, I1) -> f2(0, I1) 
5.34/5.30	  f2(I2, I3) -> f4(I2, rnd2) [rnd2 = rnd2]
5.34/5.30	  f4(I4, I5) -> f3(I4, I5) [1 + I5 <= 0]
5.34/5.30	  f4(I6, I7) -> f3(I6, I7) [1 <= I7]
5.34/5.30	  f4(I8, I9) -> f5(I8, I9) [0 <= I9 /\ I9 <= 0]
5.34/5.30	  f3(I10, I11) -> f1(0, I11) [4 <= I10 /\ I10 <= 4]
5.34/5.30	  f3(I12, I13) -> f1(I12, I13) [1 + I12 <= 4]
5.34/5.30	  f3(I14, I15) -> f1(I14, I15) [5 <= I14]
5.34/5.30	  f1(I16, I17) -> f2(1 + I16, I17) 
5.34/5.30	
5.34/5.30	EOF