3.17/3.24	MAYBE
3.17/3.24	
3.17/3.24	DP problem for innermost termination.
3.17/3.24	P =
3.17/3.24	  f10#(x1, x2) -> f1#(x1, x2) 
3.17/3.24	  f9#(I2, I3) -> f2#(I2, I3) 
3.17/3.24	  f2#(I4, I5) -> f9#(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0]
3.17/3.24	  f6#(I11, I12) -> f7#(I11, I12) [14 <= I12]
3.17/3.24	  f6#(I13, I14) -> f7#(I13, I14) [1 + I14 <= 13]
3.17/3.24	  f2#(I15, I16) -> f6#(I15, -1 + I16) 
3.17/3.24	  f5#(I17, I18) -> f2#(I17, I18) 
3.17/3.24	  f4#(I19, I20) -> f5#(I19, I20) [-1 * I20 <= 0]
3.17/3.24	  f3#(I21, I22) -> f4#(I21, I22) [14 <= I22]
3.17/3.24	  f3#(I23, I24) -> f4#(I23, I24) [1 + I24 <= 13]
3.17/3.24	  f2#(I25, I26) -> f3#(I25, -1 + I26) 
3.17/3.24	  f1#(I27, I28) -> f2#(I27, I28) 
3.17/3.24	R = 
3.17/3.24	  f10(x1, x2) -> f1(x1, x2) 
3.17/3.24	  f2(I0, I1) -> f8(rnd1, rnd2) [y1 = -1 + I1 /\ y1 <= 13 /\ 13 <= y1 /\ rnd2 = rnd2 /\ 0 <= -1 - rnd2 /\ rnd1 = rnd1]
3.17/3.24	  f9(I2, I3) -> f2(I2, I3) 
3.17/3.24	  f2(I4, I5) -> f9(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0]
3.17/3.24	  f7(I8, I9) -> f8(I10, I9) [I10 = I10 /\ 0 <= -1 - I9]
3.17/3.24	  f6(I11, I12) -> f7(I11, I12) [14 <= I12]
3.17/3.24	  f6(I13, I14) -> f7(I13, I14) [1 + I14 <= 13]
3.17/3.24	  f2(I15, I16) -> f6(I15, -1 + I16) 
3.17/3.24	  f5(I17, I18) -> f2(I17, I18) 
3.17/3.24	  f4(I19, I20) -> f5(I19, I20) [-1 * I20 <= 0]
3.17/3.24	  f3(I21, I22) -> f4(I21, I22) [14 <= I22]
3.17/3.24	  f3(I23, I24) -> f4(I23, I24) [1 + I24 <= 13]
3.17/3.24	  f2(I25, I26) -> f3(I25, -1 + I26) 
3.17/3.24	  f1(I27, I28) -> f2(I27, I28) 
3.17/3.24	
3.17/3.24	The dependency graph for this problem is:
3.17/3.24	  0 -> 11
3.17/3.24	  1 -> 2, 5, 10
3.17/3.24	  2 -> 1
3.17/3.24	  3 -> 
3.17/3.24	  4 -> 
3.17/3.24	  5 -> 3, 4
3.17/3.24	  6 -> 2, 5, 10
3.17/3.24	  7 -> 6
3.17/3.24	  8 -> 7
3.17/3.24	  9 -> 7
3.17/3.24	  10 -> 8, 9
3.17/3.24	  11 -> 2, 5, 10
3.17/3.24	Where:
3.17/3.24	  0) f10#(x1, x2) -> f1#(x1, x2) 
3.17/3.24	  1) f9#(I2, I3) -> f2#(I2, I3) 
3.17/3.24	  2) f2#(I4, I5) -> f9#(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0]
3.17/3.24	  3) f6#(I11, I12) -> f7#(I11, I12) [14 <= I12]
3.17/3.24	  4) f6#(I13, I14) -> f7#(I13, I14) [1 + I14 <= 13]
3.17/3.24	  5) f2#(I15, I16) -> f6#(I15, -1 + I16) 
3.17/3.24	  6) f5#(I17, I18) -> f2#(I17, I18) 
3.17/3.24	  7) f4#(I19, I20) -> f5#(I19, I20) [-1 * I20 <= 0]
3.17/3.24	  8) f3#(I21, I22) -> f4#(I21, I22) [14 <= I22]
3.17/3.24	  9) f3#(I23, I24) -> f4#(I23, I24) [1 + I24 <= 13]
3.17/3.24	  10) f2#(I25, I26) -> f3#(I25, -1 + I26) 
3.17/3.24	  11) f1#(I27, I28) -> f2#(I27, I28) 
3.17/3.24	
3.17/3.24	We have the following SCCs.
3.17/3.24	  { 1, 2, 6, 7, 8, 9, 10 }
3.17/3.24	
3.17/3.24	DP problem for innermost termination.
3.17/3.24	P =
3.17/3.24	  f9#(I2, I3) -> f2#(I2, I3) 
3.17/3.24	  f2#(I4, I5) -> f9#(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0]
3.17/3.24	  f5#(I17, I18) -> f2#(I17, I18) 
3.17/3.24	  f4#(I19, I20) -> f5#(I19, I20) [-1 * I20 <= 0]
3.17/3.24	  f3#(I21, I22) -> f4#(I21, I22) [14 <= I22]
3.17/3.24	  f3#(I23, I24) -> f4#(I23, I24) [1 + I24 <= 13]
3.17/3.24	  f2#(I25, I26) -> f3#(I25, -1 + I26) 
3.17/3.24	R = 
3.17/3.24	  f10(x1, x2) -> f1(x1, x2) 
3.17/3.24	  f2(I0, I1) -> f8(rnd1, rnd2) [y1 = -1 + I1 /\ y1 <= 13 /\ 13 <= y1 /\ rnd2 = rnd2 /\ 0 <= -1 - rnd2 /\ rnd1 = rnd1]
3.17/3.24	  f9(I2, I3) -> f2(I2, I3) 
3.17/3.24	  f2(I4, I5) -> f9(I4, I6) [I7 = -1 + I5 /\ I7 <= 13 /\ 13 <= I7 /\ I6 = I6 /\ -1 * I6 <= 0]
3.17/3.24	  f7(I8, I9) -> f8(I10, I9) [I10 = I10 /\ 0 <= -1 - I9]
3.17/3.24	  f6(I11, I12) -> f7(I11, I12) [14 <= I12]
3.17/3.24	  f6(I13, I14) -> f7(I13, I14) [1 + I14 <= 13]
3.17/3.24	  f2(I15, I16) -> f6(I15, -1 + I16) 
3.17/3.24	  f5(I17, I18) -> f2(I17, I18) 
3.17/3.24	  f4(I19, I20) -> f5(I19, I20) [-1 * I20 <= 0]
3.17/3.24	  f3(I21, I22) -> f4(I21, I22) [14 <= I22]
3.17/3.24	  f3(I23, I24) -> f4(I23, I24) [1 + I24 <= 13]
3.17/3.24	  f2(I25, I26) -> f3(I25, -1 + I26) 
3.17/3.24	  f1(I27, I28) -> f2(I27, I28) 
3.17/3.24	
3.17/6.21	EOF