9.54/9.41	MAYBE
9.54/9.41	
9.54/9.41	DP problem for innermost termination.
9.54/9.41	P =
9.54/9.41	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
9.54/9.41	  f7#(I0, I1, I2, I3) -> f2#(I0, I1, I0, I1) 
9.54/9.41	  f4#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5]
9.54/9.41	  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 + I8 <= I9]
9.54/9.41	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= I12]
9.54/9.41	  f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) 
9.54/9.41	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  f3#(I28, I29, I30, I31) -> f1#(I28, I29, I30, I31) [1 <= I30]
9.54/9.41	  f3#(I32, I33, I34, I35) -> f4#(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0]
9.54/9.41	  f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	R = 
9.54/9.41	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
9.54/9.41	  f7(I0, I1, I2, I3) -> f2(I0, I1, I0, I1) 
9.54/9.41	  f4(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5]
9.54/9.41	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
9.54/9.41	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= I12]
9.54/9.41	  f5(I16, I17, I18, I19) -> f6(I16, I17, I18, I19) 
9.54/9.41	  f2(I20, I21, I22, I23) -> f3(I20, I21, I22, I23) 
9.54/9.41	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  f3(I28, I29, I30, I31) -> f1(I28, I29, I30, I31) [1 <= I30]
9.54/9.41	  f3(I32, I33, I34, I35) -> f4(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0]
9.54/9.41	  f1(I36, I37, I38, I39) -> f2(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	
9.54/9.41	The dependency graph for this problem is:
9.54/9.41	  0 -> 1
9.54/9.41	  1 -> 5
9.54/9.41	  2 -> 
9.54/9.41	  3 -> 
9.54/9.41	  4 -> 
9.54/9.41	  5 -> 6, 7, 8
9.54/9.41	  6 -> 9
9.54/9.41	  7 -> 9
9.54/9.41	  8 -> 2, 3, 4
9.54/9.41	  9 -> 5
9.54/9.41	Where:
9.54/9.41	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
9.54/9.41	  1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I0, I1) 
9.54/9.41	  2) f4#(I4, I5, I6, I7) -> f5#(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5]
9.54/9.41	  3) f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 + I8 <= I9]
9.54/9.41	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= I12]
9.54/9.41	  5) f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) 
9.54/9.41	  6) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  7) f3#(I28, I29, I30, I31) -> f1#(I28, I29, I30, I31) [1 <= I30]
9.54/9.41	  8) f3#(I32, I33, I34, I35) -> f4#(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0]
9.54/9.41	  9) f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	
9.54/9.41	We have the following SCCs.
9.54/9.41	  { 5, 6, 7, 9 }
9.54/9.41	
9.54/9.41	DP problem for innermost termination.
9.54/9.41	P =
9.54/9.41	  f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) 
9.54/9.41	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  f3#(I28, I29, I30, I31) -> f1#(I28, I29, I30, I31) [1 <= I30]
9.54/9.41	  f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	R = 
9.54/9.41	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
9.54/9.41	  f7(I0, I1, I2, I3) -> f2(I0, I1, I0, I1) 
9.54/9.41	  f4(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5]
9.54/9.41	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
9.54/9.41	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= I12]
9.54/9.41	  f5(I16, I17, I18, I19) -> f6(I16, I17, I18, I19) 
9.54/9.41	  f2(I20, I21, I22, I23) -> f3(I20, I21, I22, I23) 
9.54/9.41	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  f3(I28, I29, I30, I31) -> f1(I28, I29, I30, I31) [1 <= I30]
9.54/9.41	  f3(I32, I33, I34, I35) -> f4(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0]
9.54/9.41	  f1(I36, I37, I38, I39) -> f2(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	
9.54/9.41	We use the extended value criterion with the projection function NU:
9.54/9.41	  NU[f1#(x0,x1,x2,x3)] = x2 - 2
9.54/9.41	  NU[f3#(x0,x1,x2,x3)] = x2 - 1
9.54/9.41	  NU[f2#(x0,x1,x2,x3)] = x2 - 1
9.54/9.41	
9.54/9.41	This gives the following inequalities:
9.54/9.41	    ==>  I22 - 1 >= I22 - 1
9.54/9.41	  1 + I26 <= 0  ==>  I26 - 1 >= I26 - 2
9.54/9.41	  1 <= I30  ==>  I30 - 1 > I30 - 2  with  I30 - 1 >= 0
9.54/9.41	    ==>  I38 - 2 >= (-1 + I38) - 1
9.54/9.41	
9.54/9.41	We remove all the strictly oriented dependency pairs.
9.54/9.41	
9.54/9.41	DP problem for innermost termination.
9.54/9.41	P =
9.54/9.41	  f2#(I20, I21, I22, I23) -> f3#(I20, I21, I22, I23) 
9.54/9.41	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  f1#(I36, I37, I38, I39) -> f2#(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	R = 
9.54/9.41	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
9.54/9.41	  f7(I0, I1, I2, I3) -> f2(I0, I1, I0, I1) 
9.54/9.41	  f4(I4, I5, I6, I7) -> f5(I4, I5, I6, I7) [I5 <= I4 /\ I4 <= I5]
9.54/9.41	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
9.54/9.41	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= I12]
9.54/9.41	  f5(I16, I17, I18, I19) -> f6(I16, I17, I18, I19) 
9.54/9.41	  f2(I20, I21, I22, I23) -> f3(I20, I21, I22, I23) 
9.54/9.41	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) [1 + I26 <= 0]
9.54/9.41	  f3(I28, I29, I30, I31) -> f1(I28, I29, I30, I31) [1 <= I30]
9.54/9.41	  f3(I32, I33, I34, I35) -> f4(I32, I33, I34, I35) [0 <= I34 /\ I34 <= 0]
9.54/9.41	  f1(I36, I37, I38, I39) -> f2(I36, I37, -1 + I38, -1 + I39) 
9.54/9.41	
9.57/12.39	EOF