4.13/4.13	MAYBE
4.13/4.13	
4.13/4.13	DP problem for innermost termination.
4.13/4.13	P =
4.13/4.13	  f8#(x1, x2) -> f7#(x1, x2) 
4.13/4.13	  f7#(I0, I1) -> f3#(0, I1) 
4.13/4.13	  f3#(I2, I3) -> f5#(I2, I3) 
4.13/4.13	  f5#(I4, I5) -> f4#(I4, I5) [1 + I4 <= 10]
4.13/4.13	  f4#(I8, I9) -> f2#(rnd1, I9) [rnd1 = rnd1 /\ 5 <= I8 /\ I8 <= 5]
4.13/4.13	  f4#(I10, I11) -> f1#(I10, I11) [1 + I10 <= 5]
4.13/4.13	  f4#(I12, I13) -> f1#(I12, I13) [6 <= I12]
4.13/4.13	  f2#(I14, I15) -> f3#(1 + I14, I15) 
4.13/4.13	  f1#(I16, I17) -> f2#(I16, I16) 
4.13/4.13	R = 
4.13/4.13	  f8(x1, x2) -> f7(x1, x2) 
4.13/4.13	  f7(I0, I1) -> f3(0, I1) 
4.13/4.13	  f3(I2, I3) -> f5(I2, I3) 
4.13/4.13	  f5(I4, I5) -> f4(I4, I5) [1 + I4 <= 10]
4.13/4.13	  f5(I6, I7) -> f6(I6, I7) [10 <= I6]
4.13/4.13	  f4(I8, I9) -> f2(rnd1, I9) [rnd1 = rnd1 /\ 5 <= I8 /\ I8 <= 5]
4.13/4.13	  f4(I10, I11) -> f1(I10, I11) [1 + I10 <= 5]
4.13/4.13	  f4(I12, I13) -> f1(I12, I13) [6 <= I12]
4.13/4.13	  f2(I14, I15) -> f3(1 + I14, I15) 
4.13/4.13	  f1(I16, I17) -> f2(I16, I16) 
4.13/4.13	
4.13/4.13	The dependency graph for this problem is:
4.13/4.13	  0 -> 1
4.13/4.13	  1 -> 2
4.13/4.13	  2 -> 3
4.13/4.13	  3 -> 4, 5, 6
4.13/4.13	  4 -> 7
4.13/4.13	  5 -> 8
4.13/4.13	  6 -> 8
4.13/4.13	  7 -> 2
4.13/4.13	  8 -> 7
4.13/4.13	Where:
4.13/4.13	  0) f8#(x1, x2) -> f7#(x1, x2) 
4.13/4.13	  1) f7#(I0, I1) -> f3#(0, I1) 
4.13/4.13	  2) f3#(I2, I3) -> f5#(I2, I3) 
4.13/4.13	  3) f5#(I4, I5) -> f4#(I4, I5) [1 + I4 <= 10]
4.13/4.13	  4) f4#(I8, I9) -> f2#(rnd1, I9) [rnd1 = rnd1 /\ 5 <= I8 /\ I8 <= 5]
4.13/4.13	  5) f4#(I10, I11) -> f1#(I10, I11) [1 + I10 <= 5]
4.13/4.13	  6) f4#(I12, I13) -> f1#(I12, I13) [6 <= I12]
4.13/4.13	  7) f2#(I14, I15) -> f3#(1 + I14, I15) 
4.13/4.13	  8) f1#(I16, I17) -> f2#(I16, I16) 
4.13/4.13	
4.13/4.13	We have the following SCCs.
4.13/4.13	  { 2, 3, 4, 5, 6, 7, 8 }
4.13/4.13	
4.13/4.13	DP problem for innermost termination.
4.13/4.13	P =
4.13/4.13	  f3#(I2, I3) -> f5#(I2, I3) 
4.13/4.13	  f5#(I4, I5) -> f4#(I4, I5) [1 + I4 <= 10]
4.13/4.13	  f4#(I8, I9) -> f2#(rnd1, I9) [rnd1 = rnd1 /\ 5 <= I8 /\ I8 <= 5]
4.13/4.13	  f4#(I10, I11) -> f1#(I10, I11) [1 + I10 <= 5]
4.13/4.13	  f4#(I12, I13) -> f1#(I12, I13) [6 <= I12]
4.13/4.13	  f2#(I14, I15) -> f3#(1 + I14, I15) 
4.13/4.13	  f1#(I16, I17) -> f2#(I16, I16) 
4.13/4.13	R = 
4.13/4.13	  f8(x1, x2) -> f7(x1, x2) 
4.13/4.13	  f7(I0, I1) -> f3(0, I1) 
4.13/4.13	  f3(I2, I3) -> f5(I2, I3) 
4.13/4.13	  f5(I4, I5) -> f4(I4, I5) [1 + I4 <= 10]
4.13/4.13	  f5(I6, I7) -> f6(I6, I7) [10 <= I6]
4.13/4.13	  f4(I8, I9) -> f2(rnd1, I9) [rnd1 = rnd1 /\ 5 <= I8 /\ I8 <= 5]
4.13/4.13	  f4(I10, I11) -> f1(I10, I11) [1 + I10 <= 5]
4.13/4.13	  f4(I12, I13) -> f1(I12, I13) [6 <= I12]
4.13/4.13	  f2(I14, I15) -> f3(1 + I14, I15) 
4.13/4.13	  f1(I16, I17) -> f2(I16, I16) 
4.13/4.13	
4.13/7.11	EOF