0.00/0.28	MAYBE
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  cond#(true, I0) -> f91#(f91(I0 + 11)) 
0.00/0.28	  cond#(true, I0) -> f91#(I0 + 11) 
0.00/0.28	  f91#(I1) -> cond#(I1 <= 100, I1) 
0.00/0.28	R = 
0.00/0.28	  cond(false, n) -> n - 10 
0.00/0.28	  cond(true, I0) -> f91(f91(I0 + 11)) 
0.00/0.28	  f91(I1) -> cond(I1 <= 100, I1) 
0.00/0.28	
0.00/0.28	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  cond#(true, I0) -> f91#(f91(I0 + 11)) 
0.00/0.28	  cond#(true, I0) -> f91#(I0 + 11) 
0.00/0.28	  f91#(I1) -> cond#(I1 <= 100, I1) 
0.00/0.28	  f91#(I1) -> f91#(f91(I1 + 11)) [I1 <= 100]
0.00/0.28	  f91#(I1) -> f91#(I1 + 11) [I1 <= 100]
0.00/0.28	R = 
0.00/0.28	  cond(false, n) -> n - 10 
0.00/0.28	  cond(true, I0) -> f91(f91(I0 + 11)) 
0.00/0.28	  f91(I1) -> cond(I1 <= 100, I1) 
0.00/0.28	
0.00/0.28	The dependency graph for this problem is:
0.00/0.28	  0 -> 4, 3, 2
0.00/0.28	  1 -> 4, 3, 2
0.00/0.28	  2 -> 
0.00/0.28	  3 -> 4, 3, 2
0.00/0.28	  4 -> 4, 2, 3
0.00/0.28	Where:
0.00/0.28	  0) cond#(true, I0) -> f91#(f91(I0 + 11)) 
0.00/0.28	  1) cond#(true, I0) -> f91#(I0 + 11) 
0.00/0.28	  2) f91#(I1) -> cond#(I1 <= 100, I1) 
0.00/0.28	  3) f91#(I1) -> f91#(f91(I1 + 11)) [I1 <= 100]
0.00/0.28	  4) f91#(I1) -> f91#(I1 + 11) [I1 <= 100]
0.00/0.28	
0.00/0.28	We have the following SCCs.
0.00/0.28	  { 3, 4 }
0.00/0.28	
0.00/0.28	DP problem for innermost termination.
0.00/0.28	P =
0.00/0.28	  f91#(I1) -> f91#(f91(I1 + 11)) [I1 <= 100]
0.00/0.28	  f91#(I1) -> f91#(I1 + 11) [I1 <= 100]
0.00/0.28	R = 
0.00/0.28	  cond(false, n) -> n - 10 
0.00/0.28	  cond(true, I0) -> f91(f91(I0 + 11)) 
0.00/0.28	  f91(I1) -> cond(I1 <= 100, I1) 
0.00/0.28	
0.00/3.26	EOF