2.93/3.19	MAYBE
2.93/3.19	
2.93/3.19	DP problem for innermost termination.
2.93/3.19	P =
2.93/3.19	  f6#(x1, x2) -> f5#(x1, x2) 
2.93/3.19	  f5#(I0, I1) -> f1#(I0, I1) 
2.93/3.19	  f3#(I2, I3) -> f1#(-1 + I2, 1 + I3) 
2.93/3.19	  f3#(I4, I5) -> f1#(I4, 1 + I5) 
2.93/3.19	  f4#(I6, I7) -> f2#(I6, I7) 
2.93/3.19	  f2#(I8, I9) -> f4#(I8, -1 + I9) [1 <= I9]
2.93/3.19	  f2#(I10, I11) -> f3#(I10, I11) [I11 <= 0]
2.93/3.19	  f1#(I12, I13) -> f2#(I12, I12) [1 <= I12]
2.93/3.19	R = 
2.93/3.19	  f6(x1, x2) -> f5(x1, x2) 
2.93/3.19	  f5(I0, I1) -> f1(I0, I1) 
2.93/3.19	  f3(I2, I3) -> f1(-1 + I2, 1 + I3) 
2.93/3.19	  f3(I4, I5) -> f1(I4, 1 + I5) 
2.93/3.19	  f4(I6, I7) -> f2(I6, I7) 
2.93/3.19	  f2(I8, I9) -> f4(I8, -1 + I9) [1 <= I9]
2.93/3.19	  f2(I10, I11) -> f3(I10, I11) [I11 <= 0]
2.93/3.19	  f1(I12, I13) -> f2(I12, I12) [1 <= I12]
2.93/3.19	
2.93/3.19	The dependency graph for this problem is:
2.93/3.19	  0 -> 1
2.93/3.19	  1 -> 7
2.93/3.19	  2 -> 7
2.93/3.19	  3 -> 7
2.93/3.19	  4 -> 5, 6
2.93/3.19	  5 -> 4
2.93/3.19	  6 -> 2, 3
2.93/3.19	  7 -> 5
2.93/3.19	Where:
2.93/3.19	  0) f6#(x1, x2) -> f5#(x1, x2) 
2.93/3.19	  1) f5#(I0, I1) -> f1#(I0, I1) 
2.93/3.19	  2) f3#(I2, I3) -> f1#(-1 + I2, 1 + I3) 
2.93/3.19	  3) f3#(I4, I5) -> f1#(I4, 1 + I5) 
2.93/3.19	  4) f4#(I6, I7) -> f2#(I6, I7) 
2.93/3.19	  5) f2#(I8, I9) -> f4#(I8, -1 + I9) [1 <= I9]
2.93/3.19	  6) f2#(I10, I11) -> f3#(I10, I11) [I11 <= 0]
2.93/3.19	  7) f1#(I12, I13) -> f2#(I12, I12) [1 <= I12]
2.93/3.19	
2.93/3.19	We have the following SCCs.
2.93/3.19	  { 2, 3, 4, 5, 6, 7 }
2.93/3.19	
2.93/3.19	DP problem for innermost termination.
2.93/3.19	P =
2.93/3.19	  f3#(I2, I3) -> f1#(-1 + I2, 1 + I3) 
2.93/3.19	  f3#(I4, I5) -> f1#(I4, 1 + I5) 
2.93/3.19	  f4#(I6, I7) -> f2#(I6, I7) 
2.93/3.19	  f2#(I8, I9) -> f4#(I8, -1 + I9) [1 <= I9]
2.93/3.19	  f2#(I10, I11) -> f3#(I10, I11) [I11 <= 0]
2.93/3.19	  f1#(I12, I13) -> f2#(I12, I12) [1 <= I12]
2.93/3.19	R = 
2.93/3.19	  f6(x1, x2) -> f5(x1, x2) 
2.93/3.19	  f5(I0, I1) -> f1(I0, I1) 
2.93/3.19	  f3(I2, I3) -> f1(-1 + I2, 1 + I3) 
2.93/3.19	  f3(I4, I5) -> f1(I4, 1 + I5) 
2.93/3.19	  f4(I6, I7) -> f2(I6, I7) 
2.93/3.19	  f2(I8, I9) -> f4(I8, -1 + I9) [1 <= I9]
2.93/3.19	  f2(I10, I11) -> f3(I10, I11) [I11 <= 0]
2.93/3.19	  f1(I12, I13) -> f2(I12, I12) [1 <= I12]
2.93/3.19	
2.93/6.17	EOF