1.63/1.67	YES
1.63/1.67	
1.63/1.67	DP problem for innermost termination.
1.63/1.67	P =
1.63/1.67	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
1.63/1.67	  f6#(I0, I1, I2) -> f5#(I0, I1, I2) 
1.63/1.67	  f6#(I3, I4, I5) -> f4#(I3, I4, I5) 
1.63/1.67	  f6#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.63/1.67	  f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
1.63/1.67	  f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.63/1.67	  f5#(I18, I19, I20) -> f4#(I20, I19, 0) 
1.63/1.67	  f4#(I21, I22, I23) -> f3#(I23, I22, I23) [I23 <= 19]
1.63/1.67	  f4#(I24, I25, I26) -> f1#(I26, I25, I26) [20 <= I26]
1.63/1.67	  f3#(I27, I28, I29) -> f4#(I29, I28, 1 + I29) 
1.63/1.67	R = 
1.63/1.67	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
1.63/1.67	  f6(I0, I1, I2) -> f5(I0, I1, I2) 
1.63/1.67	  f6(I3, I4, I5) -> f4(I3, I4, I5) 
1.63/1.67	  f6(I6, I7, I8) -> f3(I6, I7, I8) 
1.63/1.67	  f6(I9, I10, I11) -> f1(I9, I10, I11) 
1.63/1.67	  f6(I12, I13, I14) -> f2(I12, I13, I14) 
1.63/1.67	  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.63/1.67	  f5(I18, I19, I20) -> f4(I20, I19, 0) 
1.63/1.67	  f4(I21, I22, I23) -> f3(I23, I22, I23) [I23 <= 19]
1.63/1.67	  f4(I24, I25, I26) -> f1(I26, I25, I26) [20 <= I26]
1.63/1.67	  f3(I27, I28, I29) -> f4(I29, I28, 1 + I29) 
1.63/1.67	  f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33]
1.63/1.67	
1.63/1.67	The dependency graph for this problem is:
1.63/1.67	  0 -> 1, 2, 3, 4, 5
1.63/1.67	  1 -> 6
1.63/1.67	  2 -> 7, 8
1.63/1.67	  3 -> 9
1.63/1.67	  4 -> 
1.63/1.67	  5 -> 6
1.63/1.67	  6 -> 7
1.63/1.67	  7 -> 9
1.63/1.67	  8 -> 
1.63/1.67	  9 -> 7, 8
1.63/1.67	Where:
1.63/1.67	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
1.63/1.67	  1) f6#(I0, I1, I2) -> f5#(I0, I1, I2) 
1.63/1.67	  2) f6#(I3, I4, I5) -> f4#(I3, I4, I5) 
1.63/1.67	  3) f6#(I6, I7, I8) -> f3#(I6, I7, I8) 
1.63/1.67	  4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
1.63/1.67	  5) f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.63/1.67	  6) f5#(I18, I19, I20) -> f4#(I20, I19, 0) 
1.63/1.67	  7) f4#(I21, I22, I23) -> f3#(I23, I22, I23) [I23 <= 19]
1.63/1.67	  8) f4#(I24, I25, I26) -> f1#(I26, I25, I26) [20 <= I26]
1.63/1.67	  9) f3#(I27, I28, I29) -> f4#(I29, I28, 1 + I29) 
1.63/1.67	
1.63/1.67	We have the following SCCs.
1.63/1.67	  { 7, 9 }
1.63/1.67	
1.63/1.67	DP problem for innermost termination.
1.63/1.67	P =
1.63/1.67	  f4#(I21, I22, I23) -> f3#(I23, I22, I23) [I23 <= 19]
1.63/1.67	  f3#(I27, I28, I29) -> f4#(I29, I28, 1 + I29) 
1.63/1.67	R = 
1.63/1.67	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
1.63/1.67	  f6(I0, I1, I2) -> f5(I0, I1, I2) 
1.63/1.67	  f6(I3, I4, I5) -> f4(I3, I4, I5) 
1.63/1.67	  f6(I6, I7, I8) -> f3(I6, I7, I8) 
1.63/1.67	  f6(I9, I10, I11) -> f1(I9, I10, I11) 
1.63/1.67	  f6(I12, I13, I14) -> f2(I12, I13, I14) 
1.63/1.67	  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.63/1.67	  f5(I18, I19, I20) -> f4(I20, I19, 0) 
1.63/1.67	  f4(I21, I22, I23) -> f3(I23, I22, I23) [I23 <= 19]
1.63/1.67	  f4(I24, I25, I26) -> f1(I26, I25, I26) [20 <= I26]
1.63/1.67	  f3(I27, I28, I29) -> f4(I29, I28, 1 + I29) 
1.63/1.67	  f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33]
1.63/1.67	
1.63/1.67	We use the reverse value criterion with the projection function NU:
1.63/1.67	  NU[f3#(z1,z2,z3)] = 19 + -1 * (1 + z3)
1.63/1.67	  NU[f4#(z1,z2,z3)] = 19 + -1 * z3
1.63/1.67	
1.63/1.67	This gives the following inequalities:
1.63/1.67	  I23 <= 19  ==>  19 + -1 * I23 > 19 + -1 * (1 + I23)  with  19 + -1 * I23 >= 0
1.63/1.67	    ==>  19 + -1 * (1 + I29) >= 19 + -1 * (1 + I29)
1.63/1.67	
1.63/1.67	We remove all the strictly oriented dependency pairs.
1.63/1.67	
1.63/1.67	DP problem for innermost termination.
1.63/1.67	P =
1.63/1.67	  f3#(I27, I28, I29) -> f4#(I29, I28, 1 + I29) 
1.63/1.67	R = 
1.63/1.67	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
1.63/1.67	  f6(I0, I1, I2) -> f5(I0, I1, I2) 
1.63/1.67	  f6(I3, I4, I5) -> f4(I3, I4, I5) 
1.63/1.67	  f6(I6, I7, I8) -> f3(I6, I7, I8) 
1.63/1.67	  f6(I9, I10, I11) -> f1(I9, I10, I11) 
1.63/1.67	  f6(I12, I13, I14) -> f2(I12, I13, I14) 
1.63/1.67	  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
1.63/1.67	  f5(I18, I19, I20) -> f4(I20, I19, 0) 
1.63/1.67	  f4(I21, I22, I23) -> f3(I23, I22, I23) [I23 <= 19]
1.63/1.67	  f4(I24, I25, I26) -> f1(I26, I25, I26) [20 <= I26]
1.63/1.67	  f3(I27, I28, I29) -> f4(I29, I28, 1 + I29) 
1.63/1.67	  f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33]
1.63/1.67	
1.63/1.67	The dependency graph for this problem is:
1.63/1.67	  9 -> 
1.63/1.67	Where:
1.63/1.67	  9) f3#(I27, I28, I29) -> f4#(I29, I28, 1 + I29) 
1.63/1.67	
1.63/1.67	We have the following SCCs.
1.63/1.67	  
1.63/4.65	EOF