3.63/3.60	MAYBE
3.63/3.60	
3.63/3.60	DP problem for innermost termination.
3.63/3.60	P =
3.63/3.60	  f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
3.63/3.60	  f6#(I0, I1, I2) -> f2#(0, I1, I2) 
3.63/3.60	  f2#(I3, I4, I5) -> f4#(1, I4, I4) 
3.63/3.60	  f4#(I9, I10, I11) -> f1#(0, 1 + I10, I11) 
3.63/3.60	  f4#(I12, I13, I14) -> f1#(I12, I13, I14) [I13 <= I13]
3.63/3.60	  f1#(I18, I19, I20) -> f2#(I18, I19, I20) [1 + I20 <= I19]
3.63/3.60	R = 
3.63/3.60	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.63/3.60	  f6(I0, I1, I2) -> f2(0, I1, I2) 
3.63/3.60	  f2(I3, I4, I5) -> f4(1, I4, I4) 
3.63/3.60	  f2(I6, I7, I8) -> f5(I6, I7, I8) [1 <= I6 /\ I6 <= 1]
3.63/3.60	  f4(I9, I10, I11) -> f1(0, 1 + I10, I11) 
3.63/3.60	  f4(I12, I13, I14) -> f1(I12, I13, I14) [I13 <= I13]
3.63/3.60	  f1(I15, I16, I17) -> f3(I15, I16, I17) [I16 <= I17]
3.63/3.60	  f1(I18, I19, I20) -> f2(I18, I19, I20) [1 + I20 <= I19]
3.63/3.60	
3.63/3.60	The dependency graph for this problem is:
3.63/3.60	  0 -> 1
3.63/3.60	  1 -> 2
3.63/3.60	  2 -> 3, 4
3.63/3.60	  3 -> 5
3.63/3.60	  4 -> 5
3.63/3.60	  5 -> 2
3.63/3.60	Where:
3.63/3.60	  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
3.63/3.60	  1) f6#(I0, I1, I2) -> f2#(0, I1, I2) 
3.63/3.60	  2) f2#(I3, I4, I5) -> f4#(1, I4, I4) 
3.63/3.60	  3) f4#(I9, I10, I11) -> f1#(0, 1 + I10, I11) 
3.63/3.60	  4) f4#(I12, I13, I14) -> f1#(I12, I13, I14) [I13 <= I13]
3.63/3.60	  5) f1#(I18, I19, I20) -> f2#(I18, I19, I20) [1 + I20 <= I19]
3.63/3.60	
3.63/3.60	We have the following SCCs.
3.63/3.60	  { 2, 3, 4, 5 }
3.63/3.60	
3.63/3.60	DP problem for innermost termination.
3.63/3.60	P =
3.63/3.60	  f2#(I3, I4, I5) -> f4#(1, I4, I4) 
3.63/3.60	  f4#(I9, I10, I11) -> f1#(0, 1 + I10, I11) 
3.63/3.60	  f4#(I12, I13, I14) -> f1#(I12, I13, I14) [I13 <= I13]
3.63/3.60	  f1#(I18, I19, I20) -> f2#(I18, I19, I20) [1 + I20 <= I19]
3.63/3.60	R = 
3.63/3.60	  f7(x1, x2, x3) -> f6(x1, x2, x3) 
3.63/3.60	  f6(I0, I1, I2) -> f2(0, I1, I2) 
3.63/3.60	  f2(I3, I4, I5) -> f4(1, I4, I4) 
3.63/3.60	  f2(I6, I7, I8) -> f5(I6, I7, I8) [1 <= I6 /\ I6 <= 1]
3.63/3.60	  f4(I9, I10, I11) -> f1(0, 1 + I10, I11) 
3.63/3.60	  f4(I12, I13, I14) -> f1(I12, I13, I14) [I13 <= I13]
3.63/3.60	  f1(I15, I16, I17) -> f3(I15, I16, I17) [I16 <= I17]
3.63/3.60	  f1(I18, I19, I20) -> f2(I18, I19, I20) [1 + I20 <= I19]
3.63/3.60	
3.63/6.57	EOF