8.61/8.51	MAYBE
8.61/8.51	
8.61/8.51	DP problem for innermost termination.
8.61/8.51	P =
8.61/8.51	  f7#(x1, x2) -> f6#(x1, x2) 
8.61/8.51	  f6#(I0, I1) -> f2#(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
8.61/8.51	  f5#(I2, I3) -> f1#(I2, rnd2) [rnd2 = rnd2]
8.61/8.51	  f4#(I4, I5) -> f5#(I4, I5) [5 <= I4]
8.61/8.51	  f4#(I6, I7) -> f5#(I6, I7) [1 + I6 <= 4]
8.61/8.51	  f3#(I8, I9) -> f4#(I8, I9) [3 <= I8]
8.61/8.51	  f3#(I10, I11) -> f4#(I10, I11) [1 + I10 <= 2]
8.61/8.51	  f2#(I12, I13) -> f3#(I12, I13) [2 <= I12]
8.61/8.51	  f2#(I14, I15) -> f3#(I14, I15) [1 + I14 <= 1]
8.61/8.51	  f1#(I16, I17) -> f2#(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
8.61/8.51	  f1#(I18, I19) -> f2#(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
8.61/8.51	R = 
8.61/8.51	  f7(x1, x2) -> f6(x1, x2) 
8.61/8.51	  f6(I0, I1) -> f2(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
8.61/8.51	  f5(I2, I3) -> f1(I2, rnd2) [rnd2 = rnd2]
8.61/8.51	  f4(I4, I5) -> f5(I4, I5) [5 <= I4]
8.61/8.51	  f4(I6, I7) -> f5(I6, I7) [1 + I6 <= 4]
8.61/8.51	  f3(I8, I9) -> f4(I8, I9) [3 <= I8]
8.61/8.51	  f3(I10, I11) -> f4(I10, I11) [1 + I10 <= 2]
8.61/8.51	  f2(I12, I13) -> f3(I12, I13) [2 <= I12]
8.61/8.51	  f2(I14, I15) -> f3(I14, I15) [1 + I14 <= 1]
8.61/8.51	  f1(I16, I17) -> f2(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
8.61/8.51	  f1(I18, I19) -> f2(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
8.61/8.51	
8.61/8.51	The dependency graph for this problem is:
8.61/8.51	  0 -> 1
8.61/8.51	  1 -> 7
8.61/8.51	  2 -> 9, 10
8.61/8.51	  3 -> 2
8.61/8.51	  4 -> 2
8.61/8.51	  5 -> 3, 4
8.61/8.51	  6 -> 4
8.61/8.51	  7 -> 5
8.61/8.51	  8 -> 6
8.61/8.51	  9 -> 7, 8
8.61/8.51	  10 -> 7, 8
8.61/8.51	Where:
8.61/8.51	  0) f7#(x1, x2) -> f6#(x1, x2) 
8.61/8.51	  1) f6#(I0, I1) -> f2#(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
8.61/8.51	  2) f5#(I2, I3) -> f1#(I2, rnd2) [rnd2 = rnd2]
8.61/8.51	  3) f4#(I4, I5) -> f5#(I4, I5) [5 <= I4]
8.61/8.51	  4) f4#(I6, I7) -> f5#(I6, I7) [1 + I6 <= 4]
8.61/8.51	  5) f3#(I8, I9) -> f4#(I8, I9) [3 <= I8]
8.61/8.51	  6) f3#(I10, I11) -> f4#(I10, I11) [1 + I10 <= 2]
8.61/8.51	  7) f2#(I12, I13) -> f3#(I12, I13) [2 <= I12]
8.61/8.51	  8) f2#(I14, I15) -> f3#(I14, I15) [1 + I14 <= 1]
8.61/8.51	  9) f1#(I16, I17) -> f2#(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
8.61/8.51	  10) f1#(I18, I19) -> f2#(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
8.61/8.51	
8.61/8.51	We have the following SCCs.
8.61/8.51	  { 2, 3, 4, 5, 6, 7, 8, 9, 10 }
8.61/8.51	
8.61/8.51	DP problem for innermost termination.
8.61/8.51	P =
8.61/8.51	  f5#(I2, I3) -> f1#(I2, rnd2) [rnd2 = rnd2]
8.61/8.51	  f4#(I4, I5) -> f5#(I4, I5) [5 <= I4]
8.61/8.51	  f4#(I6, I7) -> f5#(I6, I7) [1 + I6 <= 4]
8.61/8.51	  f3#(I8, I9) -> f4#(I8, I9) [3 <= I8]
8.61/8.51	  f3#(I10, I11) -> f4#(I10, I11) [1 + I10 <= 2]
8.61/8.51	  f2#(I12, I13) -> f3#(I12, I13) [2 <= I12]
8.61/8.51	  f2#(I14, I15) -> f3#(I14, I15) [1 + I14 <= 1]
8.61/8.51	  f1#(I16, I17) -> f2#(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
8.61/8.51	  f1#(I18, I19) -> f2#(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
8.61/8.51	R = 
8.61/8.51	  f7(x1, x2) -> f6(x1, x2) 
8.61/8.51	  f6(I0, I1) -> f2(rnd1, I1) [1 <= rnd1 /\ rnd1 = rnd1]
8.61/8.51	  f5(I2, I3) -> f1(I2, rnd2) [rnd2 = rnd2]
8.61/8.51	  f4(I4, I5) -> f5(I4, I5) [5 <= I4]
8.61/8.51	  f4(I6, I7) -> f5(I6, I7) [1 + I6 <= 4]
8.61/8.51	  f3(I8, I9) -> f4(I8, I9) [3 <= I8]
8.61/8.51	  f3(I10, I11) -> f4(I10, I11) [1 + I10 <= 2]
8.61/8.51	  f2(I12, I13) -> f3(I12, I13) [2 <= I12]
8.61/8.51	  f2(I14, I15) -> f3(I14, I15) [1 + I14 <= 1]
8.61/8.51	  f1(I16, I17) -> f2(I17, I17) [I16 <= 2 * I17 /\ 2 * I17 <= I16]
8.61/8.51	  f1(I18, I19) -> f2(1 + 3 * I18, I19) [I18 <= 1 + 2 * I19 /\ 1 + 2 * I19 <= I18]
8.61/8.51	
8.61/11.49	EOF