0.00/0.06	MAYBE
0.00/0.06	
0.00/0.06	DP problem for innermost termination.
0.00/0.06	P =
0.00/0.06	  cu#(true, I0) -> cu#(I0 < exp(I0), I0 + 1) 
0.00/0.06	  cu#(true, I0) -> exp#(I0) 
0.00/0.06	R = 
0.00/0.06	  exp(x) -> 2 * x 
0.00/0.06	  cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) 
0.00/0.06	
0.00/0.06	This problem is converted using chaining, where edges between chained DPs are removed.
0.00/0.06	
0.00/0.06	DP problem for innermost termination.
0.00/0.06	P =
0.00/0.06	  cu#(true, I0) -> cu_1#(I0) 
0.00/0.06	  cu#(true, I0) -> exp#(I0) 
0.00/0.06	  cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) 
0.00/0.06	R = 
0.00/0.06	  exp(x) -> 2 * x 
0.00/0.06	  cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) 
0.00/0.06	
0.00/0.06	The dependency graph for this problem is:
0.00/0.06	  0 -> 2
0.00/0.06	  1 -> 
0.00/0.06	  2 -> 1, 0
0.00/0.06	Where:
0.00/0.06	  0) cu#(true, I0) -> cu_1#(I0) 
0.00/0.06	  1) cu#(true, I0) -> exp#(I0) 
0.00/0.06	  2) cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) 
0.00/0.06	
0.00/0.06	We have the following SCCs.
0.00/0.06	  { 0, 2 }
0.00/0.06	
0.00/0.06	DP problem for innermost termination.
0.00/0.06	P =
0.00/0.06	  cu#(true, I0) -> cu_1#(I0) 
0.00/0.06	  cu_1#(I0) -> cu#(I0 < exp(I0), I0 + 1) 
0.00/0.06	R = 
0.00/0.06	  exp(x) -> 2 * x 
0.00/0.06	  cu(true, I0) -> cu(I0 < exp(I0), I0 + 1) 
0.00/0.06	
0.00/3.04	EOF