7.63/7.89	MAYBE
7.63/7.89	
7.63/7.89	DP problem for innermost termination.
7.63/7.89	P =
7.63/7.89	  f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
7.63/7.89	  f5#(I0, I1, I2) -> f1#(I0, I1, I2) 
7.63/7.89	  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
7.63/7.89	  f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
7.63/7.89	  f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	R = 
7.63/7.89	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
7.63/7.89	  f5(I0, I1, I2) -> f1(I0, I1, I2) 
7.63/7.89	  f4(I3, I4, I5) -> f1(I3, I4, I5) 
7.63/7.89	  f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  f3(I9, I10, I11) -> f1(I9, I10, I11) 
7.63/7.89	  f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0]
7.63/7.89	
7.63/7.89	The dependency graph for this problem is:
7.63/7.89	  0 -> 1
7.63/7.89	  1 -> 3, 5
7.63/7.89	  2 -> 3, 5
7.63/7.89	  3 -> 2
7.63/7.89	  4 -> 3, 5
7.63/7.89	  5 -> 4
7.63/7.89	Where:
7.63/7.89	  0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
7.63/7.89	  1) f5#(I0, I1, I2) -> f1#(I0, I1, I2) 
7.63/7.89	  2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
7.63/7.89	  3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
7.63/7.89	  5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	
7.63/7.89	We have the following SCCs.
7.63/7.89	  { 2, 3, 4, 5 }
7.63/7.89	
7.63/7.89	DP problem for innermost termination.
7.63/7.89	P =
7.63/7.89	  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
7.63/7.89	  f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
7.63/7.89	  f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	R = 
7.63/7.89	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
7.63/7.89	  f5(I0, I1, I2) -> f1(I0, I1, I2) 
7.63/7.89	  f4(I3, I4, I5) -> f1(I3, I4, I5) 
7.63/7.89	  f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  f3(I9, I10, I11) -> f1(I9, I10, I11) 
7.63/7.89	  f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0]
7.63/7.89	
7.63/7.89	We use the reverse value criterion with the projection function NU:
7.63/7.89	  NU[f3#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0
7.63/7.89	  NU[f1#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0
7.63/7.89	  NU[f4#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0
7.63/7.89	
7.63/7.89	This gives the following inequalities:
7.63/7.89	    ==>  -1 - I4 + I5 + -1 * 0 >= -1 - I4 + I5 + -1 * 0
7.63/7.89	  0 <= -1 - I7 + I8  ==>  -1 - I7 + I8 + -1 * 0 > -1 - (1 + I7) + I8 + -1 * 0  with  -1 - I7 + I8 + -1 * 0 >= 0
7.63/7.89	    ==>  -1 - I10 + I11 + -1 * 0 >= -1 - I10 + I11 + -1 * 0
7.63/7.89	  0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0  ==>  -1 - I13 + I14 + -1 * 0 >= -1 - (1 + I13) + I14 + -1 * 0
7.63/7.89	
7.63/7.89	We remove all the strictly oriented dependency pairs.
7.63/7.89	
7.63/7.89	DP problem for innermost termination.
7.63/7.89	P =
7.63/7.89	  f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
7.63/7.89	  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
7.63/7.89	  f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	R = 
7.63/7.89	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
7.63/7.89	  f5(I0, I1, I2) -> f1(I0, I1, I2) 
7.63/7.89	  f4(I3, I4, I5) -> f1(I3, I4, I5) 
7.63/7.89	  f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  f3(I9, I10, I11) -> f1(I9, I10, I11) 
7.63/7.89	  f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0]
7.63/7.89	
7.63/7.89	The dependency graph for this problem is:
7.63/7.89	  2 -> 5
7.63/7.89	  4 -> 5
7.63/7.89	  5 -> 4
7.63/7.89	Where:
7.63/7.89	  2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 
7.63/7.89	  4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
7.63/7.89	  5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	
7.63/7.89	We have the following SCCs.
7.63/7.89	  { 4, 5 }
7.63/7.89	
7.63/7.89	DP problem for innermost termination.
7.63/7.89	P =
7.63/7.89	  f3#(I9, I10, I11) -> f1#(I9, I10, I11) 
7.63/7.89	  f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	R = 
7.63/7.89	  f6(x1, x2, x3) -> f5(x1, x2, x3) 
7.63/7.89	  f5(I0, I1, I2) -> f1(I0, I1, I2) 
7.63/7.89	  f4(I3, I4, I5) -> f1(I3, I4, I5) 
7.63/7.89	  f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
7.63/7.89	  f3(I9, I10, I11) -> f1(I9, I10, I11) 
7.63/7.89	  f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0]
7.63/7.89	  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0]
7.63/7.89	
7.63/10.87	EOF