7.89/7.84	MAYBE
7.89/7.84	
7.89/7.84	DP problem for innermost termination.
7.89/7.84	P =
7.89/7.84	  f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
7.89/7.84	  f6#(I0, I1, I2, I3) -> f1#(I0, 2, I2, I3) 
7.89/7.84	  f2#(I4, I5, I6, I7) -> f3#(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 
7.89/7.84	  f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) 
7.89/7.84	R = 
7.89/7.84	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
7.89/7.84	  f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) 
7.89/7.84	  f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
7.89/7.84	  f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) 
7.89/7.84	  f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) 
7.89/7.84	
7.89/7.84	The dependency graph for this problem is:
7.89/7.84	  0 -> 1
7.89/7.84	  1 -> 6
7.89/7.84	  2 -> 5
7.89/7.84	  3 -> 6
7.89/7.84	  4 -> 5
7.89/7.84	  5 -> 3, 4
7.89/7.84	  6 -> 2
7.89/7.84	Where:
7.89/7.84	  0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
7.89/7.84	  1) f6#(I0, I1, I2, I3) -> f1#(I0, 2, I2, I3) 
7.89/7.84	  2) f2#(I4, I5, I6, I7) -> f3#(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  3) f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 
7.89/7.84	  4) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 
7.89/7.84	  6) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) 
7.89/7.84	
7.89/7.84	We have the following SCCs.
7.89/7.84	  { 2, 3, 4, 5, 6 }
7.89/7.84	
7.89/7.84	DP problem for innermost termination.
7.89/7.84	P =
7.89/7.84	  f2#(I4, I5, I6, I7) -> f3#(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 
7.89/7.84	  f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) 
7.89/7.84	R = 
7.89/7.84	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
7.89/7.84	  f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) 
7.89/7.84	  f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
7.89/7.84	  f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) 
7.89/7.84	  f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) 
7.89/7.84	
7.89/7.84	We use the extended value criterion with the projection function NU:
7.89/7.84	  NU[f1#(x0,x1,x2,x3)] = x0 - x1
7.89/7.84	  NU[f4#(x0,x1,x2,x3)] = x0 - x1 - 1
7.89/7.84	  NU[f3#(x0,x1,x2,x3)] = x0 - x1 - 1
7.89/7.84	  NU[f2#(x0,x1,x2,x3)] = x0 - x1
7.89/7.84	
7.89/7.84	This gives the following inequalities:
7.89/7.84	  I5 <= I4  ==>  I4 - I5 > I4 - I5 - 1  with  I4 - I5 >= 0
7.89/7.84	    ==>  I12 - I13 - 1 >= I12 - (1 + I13)
7.89/7.84	  rnd4 = rnd4  ==>  I16 - I17 - 1 >= I16 - I17 - 1
7.89/7.84	    ==>  I20 - I21 - 1 >= I20 - I21 - 1
7.89/7.84	    ==>  I24 - I25 >= I24 - I25
7.89/7.84	
7.89/7.84	We remove all the strictly oriented dependency pairs.
7.89/7.84	
7.89/7.84	DP problem for innermost termination.
7.89/7.84	P =
7.89/7.84	  f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 
7.89/7.84	  f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) 
7.89/7.84	R = 
7.89/7.84	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
7.89/7.84	  f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) 
7.89/7.84	  f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
7.89/7.84	  f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) 
7.89/7.84	  f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) 
7.89/7.84	
7.89/7.84	The dependency graph for this problem is:
7.89/7.84	  3 -> 6
7.89/7.84	  4 -> 5
7.89/7.84	  5 -> 3, 4
7.89/7.84	  6 -> 
7.89/7.84	Where:
7.89/7.84	  3) f4#(I12, I13, I14, I15) -> f1#(I12, 1 + I13, I14, I15) 
7.89/7.84	  4) f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 
7.89/7.84	  6) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) 
7.89/7.84	
7.89/7.84	We have the following SCCs.
7.89/7.84	  { 4, 5 }
7.89/7.84	
7.89/7.84	DP problem for innermost termination.
7.89/7.84	P =
7.89/7.84	  f4#(I16, I17, I18, I19) -> f3#(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) 
7.89/7.84	R = 
7.89/7.84	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
7.89/7.84	  f6(I0, I1, I2, I3) -> f1(I0, 2, I2, I3) 
7.89/7.84	  f2(I4, I5, I6, I7) -> f3(I4, I5, I5, I7) [I5 <= I4]
7.89/7.84	  f2(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 + I8 <= I9]
7.89/7.84	  f4(I12, I13, I14, I15) -> f1(I12, 1 + I13, I14, I15) 
7.89/7.84	  f4(I16, I17, I18, I19) -> f3(I16, I17, -1 + I18, rnd4) [rnd4 = rnd4]
7.89/7.84	  f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) 
7.89/7.84	  f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) 
7.89/7.84	
7.89/10.82	EOF