12.29/12.33	MAYBE
12.29/12.33	
12.29/12.33	DP problem for innermost termination.
12.29/12.33	P =
12.29/12.33	  f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
12.29/12.33	  f4#(I0, I1, I2, I3, I4) -> f3#(I0, 2, I2, I3, I4) [I2 <= 3 /\ 0 <= I2 /\ I4 <= 3 /\ 0 <= I4]
12.29/12.33	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, 1 + I7, I9) [1 + 2 * I7 <= I6 + I9]
12.29/12.33	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I12, I14) [1 + I11 + I14 <= -1 + 2 * I12]
12.29/12.33	  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I17, I19) [I16 + I19 <= 2 * I17 /\ -1 + 2 * I17 <= I16 + I19]
12.29/12.33	  f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I23, I23, I24) 
12.29/12.33	  f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I28, I29) [I25 = I25]
12.29/12.33	R = 
12.29/12.33	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
12.29/12.33	  f4(I0, I1, I2, I3, I4) -> f3(I0, 2, I2, I3, I4) [I2 <= 3 /\ 0 <= I2 /\ I4 <= 3 /\ 0 <= I4]
12.29/12.33	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, 1 + I7, I9) [1 + 2 * I7 <= I6 + I9]
12.29/12.33	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I12, I14) [1 + I11 + I14 <= -1 + 2 * I12]
12.29/12.33	  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I17, I19) [I16 + I19 <= 2 * I17 /\ -1 + 2 * I17 <= I16 + I19]
12.29/12.33	  f2(I20, I21, I22, I23, I24) -> f3(I20, I21, I23, I23, I24) 
12.29/12.33	  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I25 = I25]
12.29/12.33	
12.29/12.33	The dependency graph for this problem is:
12.29/12.33	  0 -> 1
12.29/12.33	  1 -> 2, 3, 4
12.29/12.33	  2 -> 6
12.29/12.33	  3 -> 6
12.29/12.33	  4 -> 6
12.29/12.33	  5 -> 2, 3, 4
12.29/12.33	  6 -> 5
12.29/12.33	Where:
12.29/12.33	  0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
12.29/12.33	  1) f4#(I0, I1, I2, I3, I4) -> f3#(I0, 2, I2, I3, I4) [I2 <= 3 /\ 0 <= I2 /\ I4 <= 3 /\ 0 <= I4]
12.29/12.33	  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, 1 + I7, I9) [1 + 2 * I7 <= I6 + I9]
12.29/12.33	  3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I12, I14) [1 + I11 + I14 <= -1 + 2 * I12]
12.29/12.33	  4) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I17, I19) [I16 + I19 <= 2 * I17 /\ -1 + 2 * I17 <= I16 + I19]
12.29/12.33	  5) f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I23, I23, I24) 
12.29/12.33	  6) f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I28, I29) [I25 = I25]
12.29/12.33	
12.29/12.33	We have the following SCCs.
12.29/12.33	  { 2, 3, 4, 5, 6 }
12.29/12.33	
12.29/12.33	DP problem for innermost termination.
12.29/12.33	P =
12.29/12.33	  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, 1 + I7, I9) [1 + 2 * I7 <= I6 + I9]
12.29/12.33	  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, -1 + I12, I14) [1 + I11 + I14 <= -1 + 2 * I12]
12.29/12.33	  f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I17, I19) [I16 + I19 <= 2 * I17 /\ -1 + 2 * I17 <= I16 + I19]
12.29/12.33	  f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I23, I23, I24) 
12.29/12.33	  f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I28, I29) [I25 = I25]
12.29/12.33	R = 
12.29/12.33	  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
12.29/12.33	  f4(I0, I1, I2, I3, I4) -> f3(I0, 2, I2, I3, I4) [I2 <= 3 /\ 0 <= I2 /\ I4 <= 3 /\ 0 <= I4]
12.29/12.33	  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, 1 + I7, I9) [1 + 2 * I7 <= I6 + I9]
12.29/12.33	  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, -1 + I12, I14) [1 + I11 + I14 <= -1 + 2 * I12]
12.29/12.33	  f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I17, I19) [I16 + I19 <= 2 * I17 /\ -1 + 2 * I17 <= I16 + I19]
12.29/12.33	  f2(I20, I21, I22, I23, I24) -> f3(I20, I21, I23, I23, I24) 
12.29/12.33	  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I25 = I25]
12.29/12.33	
12.38/15.30	EOF