10.50/10.39	MAYBE
10.50/10.39	
10.50/10.39	DP problem for innermost termination.
10.50/10.39	P =
10.50/10.39	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
10.50/10.39	  f7#(I0, I1, I2, I3) -> f3#(I0, I1, 0, I3) 
10.50/10.39	  f3#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) 
10.50/10.39	  f5#(I8, I9, I10, I11) -> f4#(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2#(I28, I29, I30, I31) -> f3#(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I34) 
10.50/10.39	R = 
10.50/10.39	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
10.50/10.39	  f7(I0, I1, I2, I3) -> f3(I0, I1, 0, I3) 
10.50/10.39	  f3(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) 
10.50/10.39	  f5(I8, I9, I10, I11) -> f4(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f5(I12, I13, I14, I15) -> f6(I12, I13, I14, I15) [I12 <= I14]
10.50/10.39	  f4(I16, I17, I18, I19) -> f2(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2(I28, I29, I30, I31) -> f3(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I34) 
10.50/10.39	
10.50/10.39	The dependency graph for this problem is:
10.50/10.39	  0 -> 1
10.50/10.39	  1 -> 2
10.50/10.39	  2 -> 3
10.50/10.39	  3 -> 4, 5, 6
10.50/10.39	  4 -> 7
10.50/10.39	  5 -> 8
10.50/10.39	  6 -> 8
10.50/10.39	  7 -> 2
10.50/10.39	  8 -> 7
10.50/10.39	Where:
10.50/10.39	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
10.50/10.39	  1) f7#(I0, I1, I2, I3) -> f3#(I0, I1, 0, I3) 
10.50/10.39	  2) f3#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) 
10.50/10.39	  3) f5#(I8, I9, I10, I11) -> f4#(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  4) f4#(I16, I17, I18, I19) -> f2#(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  5) f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  6) f4#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  7) f2#(I28, I29, I30, I31) -> f3#(I28, I29, 1 + I30, I31) 
10.50/10.39	  8) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I34) 
10.50/10.39	
10.50/10.39	We have the following SCCs.
10.50/10.39	  { 2, 3, 4, 5, 6, 7, 8 }
10.50/10.39	
10.50/10.39	DP problem for innermost termination.
10.50/10.39	P =
10.50/10.39	  f3#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) 
10.50/10.39	  f5#(I8, I9, I10, I11) -> f4#(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2#(I28, I29, I30, I31) -> f3#(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I34) 
10.50/10.39	R = 
10.50/10.39	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
10.50/10.39	  f7(I0, I1, I2, I3) -> f3(I0, I1, 0, I3) 
10.50/10.39	  f3(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) 
10.50/10.39	  f5(I8, I9, I10, I11) -> f4(I8, I9, I10, I11) [1 + I10 <= I8]
10.50/10.39	  f5(I12, I13, I14, I15) -> f6(I12, I13, I14, I15) [I12 <= I14]
10.50/10.39	  f4(I16, I17, I18, I19) -> f2(I16, I17, rnd3, I19) [rnd3 = rnd3 /\ I17 <= I18 /\ I18 <= I17]
10.50/10.39	  f4(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) [1 + I22 <= I21]
10.50/10.39	  f4(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I25 <= I26]
10.50/10.39	  f2(I28, I29, I30, I31) -> f3(I28, I29, 1 + I30, I31) 
10.50/10.39	  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I34) 
10.50/10.39	
10.50/13.37	EOF