9.42/9.35	MAYBE
9.42/9.35	
9.42/9.35	DP problem for innermost termination.
9.42/9.35	P =
9.42/9.35	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
9.42/9.35	  f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
9.42/9.35	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
9.42/9.35	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
9.42/9.35	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
9.42/9.35	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
9.42/9.35	  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
9.42/9.35	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
9.42/9.35	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, 1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
9.42/9.35	R = 
9.42/9.35	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
9.42/9.35	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
9.42/9.35	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
9.42/9.35	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
9.42/9.35	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
9.42/9.35	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
9.42/9.35	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
9.42/9.35	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
9.42/9.35	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, 1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
9.42/9.35	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
9.42/9.35	
9.42/9.35	The dependency graph for this problem is:
9.42/9.35	  0 -> 1
9.42/9.35	  1 -> 6, 8
9.42/9.35	  2 -> 6, 8
9.42/9.35	  3 -> 2
9.42/9.35	  4 -> 3
9.42/9.35	  5 -> 3
9.42/9.35	  6 -> 4, 5
9.42/9.35	  7 -> 6, 8
9.42/9.35	  8 -> 7
9.42/9.35	Where:
9.42/9.35	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
9.42/9.35	  1) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
9.42/9.35	  2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
9.42/9.35	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
9.42/9.35	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
9.42/9.35	  5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
9.42/9.35	  6) f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
9.42/9.35	  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
9.42/9.35	  8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, 1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
9.42/9.35	
9.42/9.35	We have the following SCCs.
9.42/9.35	  { 2, 3, 4, 5, 6, 7, 8 }
9.42/9.35	
9.42/9.35	DP problem for innermost termination.
9.42/9.35	P =
9.42/9.35	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
9.42/9.35	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
9.42/9.35	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
9.42/9.35	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
9.42/9.35	  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
9.42/9.35	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
9.42/9.35	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, 1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
9.42/9.35	R = 
9.42/9.35	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
9.42/9.35	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
9.42/9.35	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
9.42/9.35	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
9.42/9.35	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
9.42/9.35	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
9.42/9.35	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
9.42/9.35	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
9.42/9.35	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, 1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
9.42/9.35	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
9.42/9.35	
9.42/12.32	EOF