9.12/9.15	MAYBE
9.12/9.15	
9.12/9.15	DP problem for innermost termination.
9.12/9.15	P =
9.12/9.15	  f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
9.12/9.15	  f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.15	  f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 
9.12/9.15	  f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, 1 + I25) 
9.12/9.15	  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.15	  f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.15	R = 
9.12/9.15	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
9.12/9.15	  f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0]
9.12/9.15	  f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.15	  f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 
9.12/9.15	  f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0]
9.12/9.15	  f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, 1 + I25) 
9.12/9.15	  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.15	  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.15	
9.12/9.15	The dependency graph for this problem is:
9.12/9.15	  0 -> 2
9.12/9.15	  1 -> 5
9.12/9.15	  2 -> 1
9.12/9.15	  3 -> 1
9.12/9.15	  4 -> 3
9.12/9.15	  5 -> 4
9.12/9.15	Where:
9.12/9.15	  0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 
9.12/9.15	  1) f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.15	  2) f6#(I10, I11, I12, I13, I14) -> f4#(I10, I11, 0, I13, I14) 
9.12/9.15	  3) f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, 1 + I25) 
9.12/9.15	  4) f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.15	  5) f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.15	
9.12/9.15	We have the following SCCs.
9.12/9.15	  { 1, 3, 4, 5 }
9.12/9.15	
9.12/9.15	DP problem for innermost termination.
9.12/9.15	P =
9.12/9.15	  f4#(I5, I6, I7, I8, I9) -> f1#(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.15	  f3#(I21, I22, I23, I24, I25) -> f4#(I21, I22, 0, I24, 1 + I25) 
9.12/9.15	  f2#(I26, I27, I28, I29, I30) -> f3#(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.15	  f1#(I31, I32, I33, I34, I35) -> f2#(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.15	R = 
9.12/9.15	  f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 
9.12/9.15	  f4(I0, I1, I2, I3, I4) -> f5(rnd1, I1, I2, I3, I4) [rnd1 = rnd1 /\ -1 * I3 + I4 <= 0]
9.12/9.15	  f4(I5, I6, I7, I8, I9) -> f1(I5, I6, 1, 1 + I8, I9) [0 <= I7 /\ I7 <= 0 /\ 0 <= -1 - I8 + I9]
9.12/9.15	  f6(I10, I11, I12, I13, I14) -> f4(I10, I11, 0, I13, I14) 
9.12/9.15	  f1(I15, I16, I17, I18, I19) -> f5(I20, I16, I17, I18, I19) [I20 = I20 /\ -1 * I18 + I19 <= 0]
9.12/9.15	  f3(I21, I22, I23, I24, I25) -> f4(I21, I22, 0, I24, 1 + I25) 
9.12/9.15	  f2(I26, I27, I28, I29, I30) -> f3(I26, I27, I28, I29, I30) [I27 = I27]
9.12/9.15	  f1(I31, I32, I33, I34, I35) -> f2(I31, I32, I33, I34, I35) [0 <= -1 - I34 + I35]
9.12/9.15	
9.16/12.12	EOF