10.23/10.11	MAYBE
10.23/10.11	
10.23/10.11	DP problem for innermost termination.
10.23/10.11	P =
10.23/10.11	  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
10.23/10.11	  f4#(I0, I1, I2, I3) -> f3#(2, I1, I2, I3) [I1 <= 3 /\ 0 <= I1 /\ I3 <= 3 /\ 0 <= I3]
10.23/10.11	  f3#(I4, I5, I6, I7) -> f1#(I4, I5, 1 + I5, I7) [1 + 2 * I5 <= I4 + I7]
10.23/10.11	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, -1 + I9, I11) [1 + I8 + I11 <= -1 + 2 * I9]
10.23/10.11	  f3#(I12, I13, I14, I15) -> f1#(I12, I13, I13, I15) [I12 + I15 <= 2 * I13 /\ -1 + 2 * I13 <= I12 + I15]
10.23/10.11	  f2#(I16, I17, I18, I19) -> f3#(I16, I18, I18, I19) 
10.23/10.11	  f1#(I20, I21, I22, I23) -> f2#(I20, I21, I22, I23) [1 + I22 <= I21]
10.23/10.11	  f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) [1 + I25 <= I26]
10.23/10.11	R = 
10.23/10.11	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
10.23/10.11	  f4(I0, I1, I2, I3) -> f3(2, I1, I2, I3) [I1 <= 3 /\ 0 <= I1 /\ I3 <= 3 /\ 0 <= I3]
10.23/10.11	  f3(I4, I5, I6, I7) -> f1(I4, I5, 1 + I5, I7) [1 + 2 * I5 <= I4 + I7]
10.23/10.11	  f3(I8, I9, I10, I11) -> f1(I8, I9, -1 + I9, I11) [1 + I8 + I11 <= -1 + 2 * I9]
10.23/10.11	  f3(I12, I13, I14, I15) -> f1(I12, I13, I13, I15) [I12 + I15 <= 2 * I13 /\ -1 + 2 * I13 <= I12 + I15]
10.23/10.11	  f2(I16, I17, I18, I19) -> f3(I16, I18, I18, I19) 
10.23/10.11	  f1(I20, I21, I22, I23) -> f2(I20, I21, I22, I23) [1 + I22 <= I21]
10.23/10.11	  f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) [1 + I25 <= I26]
10.23/10.11	
10.23/10.11	The dependency graph for this problem is:
10.23/10.11	  0 -> 1
10.23/10.11	  1 -> 2, 3, 4
10.23/10.11	  2 -> 7
10.23/10.11	  3 -> 6
10.23/10.11	  4 -> 
10.23/10.11	  5 -> 2, 3, 4
10.23/10.11	  6 -> 5
10.23/10.11	  7 -> 5
10.23/10.11	Where:
10.23/10.11	  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
10.23/10.11	  1) f4#(I0, I1, I2, I3) -> f3#(2, I1, I2, I3) [I1 <= 3 /\ 0 <= I1 /\ I3 <= 3 /\ 0 <= I3]
10.23/10.11	  2) f3#(I4, I5, I6, I7) -> f1#(I4, I5, 1 + I5, I7) [1 + 2 * I5 <= I4 + I7]
10.23/10.11	  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, -1 + I9, I11) [1 + I8 + I11 <= -1 + 2 * I9]
10.23/10.11	  4) f3#(I12, I13, I14, I15) -> f1#(I12, I13, I13, I15) [I12 + I15 <= 2 * I13 /\ -1 + 2 * I13 <= I12 + I15]
10.23/10.11	  5) f2#(I16, I17, I18, I19) -> f3#(I16, I18, I18, I19) 
10.23/10.11	  6) f1#(I20, I21, I22, I23) -> f2#(I20, I21, I22, I23) [1 + I22 <= I21]
10.23/10.11	  7) f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) [1 + I25 <= I26]
10.23/10.11	
10.23/10.11	We have the following SCCs.
10.23/10.11	  { 2, 3, 5, 6, 7 }
10.23/10.11	
10.23/10.11	DP problem for innermost termination.
10.23/10.11	P =
10.23/10.11	  f3#(I4, I5, I6, I7) -> f1#(I4, I5, 1 + I5, I7) [1 + 2 * I5 <= I4 + I7]
10.23/10.11	  f3#(I8, I9, I10, I11) -> f1#(I8, I9, -1 + I9, I11) [1 + I8 + I11 <= -1 + 2 * I9]
10.23/10.11	  f2#(I16, I17, I18, I19) -> f3#(I16, I18, I18, I19) 
10.23/10.11	  f1#(I20, I21, I22, I23) -> f2#(I20, I21, I22, I23) [1 + I22 <= I21]
10.23/10.11	  f1#(I24, I25, I26, I27) -> f2#(I24, I25, I26, I27) [1 + I25 <= I26]
10.23/10.11	R = 
10.23/10.11	  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
10.23/10.11	  f4(I0, I1, I2, I3) -> f3(2, I1, I2, I3) [I1 <= 3 /\ 0 <= I1 /\ I3 <= 3 /\ 0 <= I3]
10.23/10.11	  f3(I4, I5, I6, I7) -> f1(I4, I5, 1 + I5, I7) [1 + 2 * I5 <= I4 + I7]
10.23/10.11	  f3(I8, I9, I10, I11) -> f1(I8, I9, -1 + I9, I11) [1 + I8 + I11 <= -1 + 2 * I9]
10.23/10.11	  f3(I12, I13, I14, I15) -> f1(I12, I13, I13, I15) [I12 + I15 <= 2 * I13 /\ -1 + 2 * I13 <= I12 + I15]
10.23/10.11	  f2(I16, I17, I18, I19) -> f3(I16, I18, I18, I19) 
10.23/10.11	  f1(I20, I21, I22, I23) -> f2(I20, I21, I22, I23) [1 + I22 <= I21]
10.23/10.11	  f1(I24, I25, I26, I27) -> f2(I24, I25, I26, I27) [1 + I25 <= I26]
10.23/10.11	
10.23/13.09	EOF