15.85/15.64	MAYBE
15.85/15.64	
15.85/15.64	DP problem for innermost termination.
15.85/15.64	P =
15.85/15.64	  f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
15.85/15.64	  f6#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 
15.85/15.64	  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.85/15.64	  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
15.85/15.64	  f1#(I25, I26, I27, I28, I29) -> f3#(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	R = 
15.85/15.64	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
15.85/15.64	  f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.85/15.64	  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.85/15.64	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
15.85/15.64	  f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	  f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0]
15.85/15.64	
15.85/15.64	The dependency graph for this problem is:
15.85/15.64	  0 -> 1
15.85/15.64	  1 -> 4, 6
15.85/15.64	  2 -> 4, 6
15.85/15.64	  3 -> 2
15.85/15.64	  4 -> 3
15.85/15.64	  5 -> 4, 6
15.85/15.64	  6 -> 5
15.85/15.64	Where:
15.85/15.64	  0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
15.85/15.64	  1) f6#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 
15.85/15.64	  2) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.85/15.64	  3) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  4) f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  5) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
15.85/15.64	  6) f1#(I25, I26, I27, I28, I29) -> f3#(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	
15.85/15.64	We have the following SCCs.
15.85/15.64	  { 2, 3, 4, 5, 6 }
15.85/15.64	
15.85/15.64	DP problem for innermost termination.
15.85/15.64	P =
15.85/15.64	  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.85/15.64	  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
15.85/15.64	  f1#(I25, I26, I27, I28, I29) -> f3#(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	R = 
15.85/15.64	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
15.85/15.64	  f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.85/15.64	  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.85/15.64	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
15.85/15.64	  f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	  f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0]
15.85/15.64	
15.85/15.64	We use the extended value criterion with the projection function NU:
15.85/15.64	  NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.85/15.64	  NU[f4#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.85/15.64	  NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.85/15.64	  NU[f5#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1
15.85/15.64	
15.85/15.64	This gives the following inequalities:
15.85/15.64	    ==>  -I8 + I9 - 1 >= -I8 + I9 - 1
15.85/15.64	  I11 = I11  ==>  -I13 + I14 - 1 >= -I13 + I14 - 1
15.85/15.64	  rnd3 = rnd3 /\ 0 <= -1 - I18 + I19  ==>  -I18 + I19 - 1 >= -I18 + I19 - 1
15.85/15.64	    ==>  -I23 + I24 - 1 >= -I23 + I24 - 1
15.85/15.64	  0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29  ==>  -I28 + I29 - 1 > -I28 + (-1 + I29) - 1  with  -I28 + I29 - 1 >= 0
15.85/15.64	
15.85/15.64	We remove all the strictly oriented dependency pairs.
15.85/15.64	
15.85/15.64	DP problem for innermost termination.
15.85/15.64	P =
15.85/15.64	  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.85/15.64	  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
15.85/15.64	R = 
15.85/15.64	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
15.85/15.64	  f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.85/15.64	  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.85/15.64	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
15.85/15.64	  f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	  f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0]
15.85/15.64	
15.85/15.64	The dependency graph for this problem is:
15.85/15.64	  2 -> 4
15.85/15.64	  3 -> 2
15.85/15.64	  4 -> 3
15.85/15.64	  5 -> 4
15.85/15.64	Where:
15.85/15.64	  2) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.85/15.64	  3) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  4) f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  5) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 
15.85/15.64	
15.85/15.64	We have the following SCCs.
15.85/15.64	  { 2, 3, 4 }
15.85/15.64	
15.85/15.64	DP problem for innermost termination.
15.85/15.64	P =
15.85/15.64	  f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
15.85/15.64	  f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	R = 
15.85/15.64	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
15.85/15.64	  f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) 
15.85/15.64	  f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
15.85/15.64	  f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11]
15.85/15.64	  f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19]
15.85/15.64	  f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) 
15.85/15.64	  f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29]
15.85/15.64	  f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0]
15.85/15.64	
15.85/18.61	EOF