11.04/10.92	MAYBE
11.04/10.92	
11.04/10.92	DP problem for innermost termination.
11.04/10.92	P =
11.04/10.92	  f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
11.04/10.92	  f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
11.04/10.92	  f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 
11.04/10.92	  f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	  f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) 
11.04/10.92	R = 
11.04/10.92	  f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
11.04/10.92	  f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
11.04/10.92	  f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
11.04/10.92	  f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	  f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28]
11.04/10.92	  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 
11.04/10.92	
11.04/10.92	The dependency graph for this problem is:
11.04/10.92	  0 -> 7
11.04/10.92	  1 -> 4, 6
11.04/10.92	  2 -> 1
11.04/10.92	  3 -> 2
11.04/10.92	  4 -> 3
11.04/10.92	  5 -> 4, 6
11.04/10.92	  6 -> 5
11.04/10.92	  7 -> 4, 6
11.04/10.92	Where:
11.04/10.92	  0) f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
11.04/10.92	  1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
11.04/10.92	  2) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 
11.04/10.92	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  4) f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  5) f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 
11.04/10.92	  6) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	  7) f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) 
11.04/10.92	
11.04/10.92	We have the following SCCs.
11.04/10.92	  { 1, 2, 3, 4, 5, 6 }
11.04/10.92	
11.04/10.92	DP problem for innermost termination.
11.04/10.92	P =
11.04/10.92	  f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
11.04/10.92	  f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 
11.04/10.92	  f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	R = 
11.04/10.92	  f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
11.04/10.92	  f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
11.04/10.92	  f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
11.04/10.92	  f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	  f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28]
11.04/10.92	  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 
11.04/10.92	
11.04/10.92	We use the extended value criterion with the projection function NU:
11.04/10.92	  NU[f4#(x0,x1,x2,x3)] = x3
11.04/10.92	  NU[f5#(x0,x1,x2,x3)] = x3 - 1
11.04/10.92	  NU[f6#(x0,x1,x2,x3)] = x3 - 1
11.04/10.92	  NU[f2#(x0,x1,x2,x3)] = x3
11.04/10.92	  NU[f7#(x0,x1,x2,x3)] = x3
11.04/10.92	
11.04/10.92	This gives the following inequalities:
11.04/10.92	    ==>  I3 >= I3
11.04/10.92	    ==>  I7 - 1 >= (-1 + I7)
11.04/10.92	  I9 = I9  ==>  I11 - 1 >= I11 - 1
11.04/10.92	  rnd3 = rnd3 /\ -1 * I15 <= 0  ==>  I15 > I15 - 1  with  I15 >= 0
11.04/10.92	    ==>  I19 >= I19
11.04/10.92	  0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0  ==>  I23 >= I23
11.04/10.92	
11.04/10.92	We remove all the strictly oriented dependency pairs.
11.04/10.92	
11.04/10.92	DP problem for innermost termination.
11.04/10.92	P =
11.04/10.92	  f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
11.04/10.92	  f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 
11.04/10.92	  f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	R = 
11.04/10.92	  f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
11.04/10.92	  f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
11.04/10.92	  f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
11.04/10.92	  f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	  f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28]
11.04/10.92	  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 
11.04/10.92	
11.04/10.92	The dependency graph for this problem is:
11.04/10.92	  1 -> 6
11.04/10.92	  2 -> 1
11.04/10.92	  3 -> 2
11.04/10.92	  5 -> 6
11.04/10.92	  6 -> 5
11.04/10.92	Where:
11.04/10.92	  1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
11.04/10.92	  2) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 
11.04/10.92	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  5) f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 
11.04/10.92	  6) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	
11.04/10.92	We have the following SCCs.
11.04/10.92	  { 5, 6 }
11.04/10.92	
11.04/10.92	DP problem for innermost termination.
11.04/10.92	P =
11.04/10.92	  f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 
11.04/10.92	  f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	R = 
11.04/10.92	  f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
11.04/10.92	  f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
11.04/10.92	  f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 
11.04/10.92	  f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9]
11.04/10.92	  f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0]
11.04/10.92	  f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 
11.04/10.92	  f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0]
11.04/10.92	  f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28]
11.04/10.92	  f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 
11.04/10.92	
11.04/10.93	EOF