11.54/11.40	MAYBE
11.54/11.40	
11.54/11.40	DP problem for innermost termination.
11.54/11.40	P =
11.54/11.40	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
11.54/11.40	  f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
11.54/11.40	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
11.54/11.40	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
11.54/11.40	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	R = 
11.54/11.40	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
11.54/11.40	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
11.54/11.40	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
11.54/11.40	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
11.54/11.40	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
11.54/11.40	
11.54/11.40	The dependency graph for this problem is:
11.54/11.40	  0 -> 1
11.54/11.40	  1 -> 6, 8
11.54/11.40	  2 -> 6, 8
11.54/11.40	  3 -> 2
11.54/11.40	  4 -> 3
11.54/11.40	  5 -> 3
11.54/11.40	  6 -> 4, 5
11.54/11.40	  7 -> 6, 8
11.54/11.40	  8 -> 7
11.54/11.40	Where:
11.54/11.40	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
11.54/11.40	  1) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
11.54/11.40	  2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
11.54/11.40	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
11.54/11.40	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  6) f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
11.54/11.40	  8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	
11.54/11.40	We have the following SCCs.
11.54/11.40	  { 2, 3, 4, 5, 6, 7, 8 }
11.54/11.40	
11.54/11.40	DP problem for innermost termination.
11.54/11.40	P =
11.54/11.40	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
11.54/11.40	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
11.54/11.40	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	R = 
11.54/11.40	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
11.54/11.40	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
11.54/11.40	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
11.54/11.40	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
11.54/11.40	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
11.54/11.40	
11.54/11.40	We use the extended value criterion with the projection function NU:
11.54/11.40	  NU[f3#(x0,x1,x2,x3)] = -x2 + x3 + 1
11.54/11.40	  NU[f4#(x0,x1,x2,x3)] = -x2 + x3
11.54/11.40	  NU[f5#(x0,x1,x2,x3)] = -x2 + x3
11.54/11.40	  NU[f1#(x0,x1,x2,x3)] = -x2 + x3 + 1
11.54/11.40	  NU[f6#(x0,x1,x2,x3)] = -x2 + x3 + 1
11.54/11.40	
11.54/11.40	This gives the following inequalities:
11.54/11.40	    ==>  -I6 + I7 + 1 >= -I6 + I7 + 1
11.54/11.40	    ==>  -I10 + I11 >= -(1 + I10) + I11 + 1
11.54/11.40	  1 <= I13  ==>  -I14 + I15 >= -I14 + I15
11.54/11.40	  1 + I17 <= 0  ==>  -I18 + I19 >= -I18 + I19
11.54/11.40	  rnd2 = rnd2 /\ 0 <= -1 - I22 + I23  ==>  -I22 + I23 + 1 > -I22 + I23  with  -I22 + I23 + 1 >= 0
11.54/11.40	    ==>  -I26 + I27 + 1 >= -I26 + I27 + 1
11.54/11.40	  0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31  ==>  -I30 + I31 + 1 >= -I30 + I31 + 1
11.54/11.40	
11.54/11.40	We remove all the strictly oriented dependency pairs.
11.54/11.40	
11.54/11.40	DP problem for innermost termination.
11.54/11.40	P =
11.54/11.40	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
11.54/11.40	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
11.54/11.40	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	R = 
11.54/11.40	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
11.54/11.40	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
11.54/11.40	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
11.54/11.40	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
11.54/11.40	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
11.54/11.40	
11.54/11.40	The dependency graph for this problem is:
11.54/11.40	  2 -> 8
11.54/11.40	  3 -> 2
11.54/11.40	  4 -> 3
11.54/11.40	  5 -> 3
11.54/11.40	  7 -> 8
11.54/11.40	  8 -> 7
11.54/11.40	Where:
11.54/11.40	  2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
11.54/11.40	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
11.54/11.40	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
11.54/11.40	  8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	
11.54/11.40	We have the following SCCs.
11.54/11.40	  { 7, 8 }
11.54/11.40	
11.54/11.40	DP problem for innermost termination.
11.54/11.40	P =
11.54/11.40	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
11.54/11.40	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	R = 
11.54/11.40	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
11.54/11.40	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
11.54/11.40	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
11.54/11.40	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
11.54/11.40	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
11.54/11.40	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
11.54/11.40	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
11.54/11.40	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
11.54/11.40	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
11.54/11.40	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
11.54/11.40	
11.54/11.40	EOF