1.03/1.08	MAYBE
1.03/1.08	
1.03/1.08	DP problem for innermost termination.
1.03/1.08	P =
1.03/1.08	  f10#(x1, x2) -> f9#(x1, x2) 
1.03/1.08	  f9#(I0, I1) -> f2#(0, rnd2) [rnd2 = rnd2]
1.03/1.08	  f4#(I2, I3) -> f8#(I2, I3) [0 <= I3]
1.03/1.08	  f4#(I4, I5) -> f5#(I4, I5) [1 + I5 <= 0]
1.03/1.08	  f8#(I6, I7) -> f7#(I6, I7) [1 + I7 <= I6]
1.03/1.08	  f8#(I8, I9) -> f5#(I8, I9) [I8 <= I9]
1.03/1.08	  f7#(I12, I13) -> f5#(I12, I13) 
1.03/1.08	  f7#(I14, I15) -> f5#(I14, I15) 
1.03/1.08	  f2#(I18, I19) -> f3#(I18, I19) 
1.03/1.08	  f3#(I20, I21) -> f4#(I20, I21) 
1.03/1.08	  f3#(I22, I23) -> f1#(I22, I23) 
1.03/1.08	  f3#(I24, I25) -> f1#(I24, I25) 
1.03/1.08	  f1#(I26, I27) -> f2#(1 + I26, I27) 
1.03/1.08	R = 
1.03/1.08	  f10(x1, x2) -> f9(x1, x2) 
1.03/1.08	  f9(I0, I1) -> f2(0, rnd2) [rnd2 = rnd2]
1.03/1.08	  f4(I2, I3) -> f8(I2, I3) [0 <= I3]
1.03/1.08	  f4(I4, I5) -> f5(I4, I5) [1 + I5 <= 0]
1.03/1.08	  f8(I6, I7) -> f7(I6, I7) [1 + I7 <= I6]
1.03/1.08	  f8(I8, I9) -> f5(I8, I9) [I8 <= I9]
1.03/1.08	  f7(I10, I11) -> f6(I10, I11) 
1.03/1.08	  f7(I12, I13) -> f5(I12, I13) 
1.03/1.08	  f7(I14, I15) -> f5(I14, I15) 
1.03/1.08	  f5(I16, I17) -> f6(I16, I17) 
1.03/1.08	  f2(I18, I19) -> f3(I18, I19) 
1.03/1.08	  f3(I20, I21) -> f4(I20, I21) 
1.03/1.08	  f3(I22, I23) -> f1(I22, I23) 
1.03/1.08	  f3(I24, I25) -> f1(I24, I25) 
1.03/1.08	  f1(I26, I27) -> f2(1 + I26, I27) 
1.03/1.08	
1.03/1.08	The dependency graph for this problem is:
1.03/1.08	  0 -> 1
1.03/1.08	  1 -> 8
1.03/1.08	  2 -> 4, 5
1.03/1.08	  3 -> 
1.03/1.08	  4 -> 6, 7
1.03/1.08	  5 -> 
1.03/1.08	  6 -> 
1.03/1.08	  7 -> 
1.03/1.08	  8 -> 9, 10, 11
1.03/1.08	  9 -> 2, 3
1.03/1.08	  10 -> 12
1.03/1.08	  11 -> 12
1.03/1.08	  12 -> 8
1.03/1.08	Where:
1.03/1.08	  0) f10#(x1, x2) -> f9#(x1, x2) 
1.03/1.08	  1) f9#(I0, I1) -> f2#(0, rnd2) [rnd2 = rnd2]
1.03/1.08	  2) f4#(I2, I3) -> f8#(I2, I3) [0 <= I3]
1.03/1.08	  3) f4#(I4, I5) -> f5#(I4, I5) [1 + I5 <= 0]
1.03/1.08	  4) f8#(I6, I7) -> f7#(I6, I7) [1 + I7 <= I6]
1.03/1.08	  5) f8#(I8, I9) -> f5#(I8, I9) [I8 <= I9]
1.03/1.08	  6) f7#(I12, I13) -> f5#(I12, I13) 
1.03/1.08	  7) f7#(I14, I15) -> f5#(I14, I15) 
1.03/1.08	  8) f2#(I18, I19) -> f3#(I18, I19) 
1.03/1.08	  9) f3#(I20, I21) -> f4#(I20, I21) 
1.03/1.08	  10) f3#(I22, I23) -> f1#(I22, I23) 
1.03/1.08	  11) f3#(I24, I25) -> f1#(I24, I25) 
1.03/1.08	  12) f1#(I26, I27) -> f2#(1 + I26, I27) 
1.03/1.08	
1.03/1.08	We have the following SCCs.
1.03/1.08	  { 8, 10, 11, 12 }
1.03/1.08	
1.03/1.08	DP problem for innermost termination.
1.03/1.08	P =
1.03/1.08	  f2#(I18, I19) -> f3#(I18, I19) 
1.03/1.08	  f3#(I22, I23) -> f1#(I22, I23) 
1.03/1.08	  f3#(I24, I25) -> f1#(I24, I25) 
1.03/1.08	  f1#(I26, I27) -> f2#(1 + I26, I27) 
1.03/1.08	R = 
1.03/1.08	  f10(x1, x2) -> f9(x1, x2) 
1.03/1.08	  f9(I0, I1) -> f2(0, rnd2) [rnd2 = rnd2]
1.03/1.08	  f4(I2, I3) -> f8(I2, I3) [0 <= I3]
1.03/1.08	  f4(I4, I5) -> f5(I4, I5) [1 + I5 <= 0]
1.03/1.08	  f8(I6, I7) -> f7(I6, I7) [1 + I7 <= I6]
1.03/1.08	  f8(I8, I9) -> f5(I8, I9) [I8 <= I9]
1.03/1.08	  f7(I10, I11) -> f6(I10, I11) 
1.03/1.08	  f7(I12, I13) -> f5(I12, I13) 
1.03/1.08	  f7(I14, I15) -> f5(I14, I15) 
1.03/1.08	  f5(I16, I17) -> f6(I16, I17) 
1.03/1.08	  f2(I18, I19) -> f3(I18, I19) 
1.03/1.08	  f3(I20, I21) -> f4(I20, I21) 
1.03/1.08	  f3(I22, I23) -> f1(I22, I23) 
1.03/1.08	  f3(I24, I25) -> f1(I24, I25) 
1.03/1.08	  f1(I26, I27) -> f2(1 + I26, I27) 
1.03/1.08	
1.03/4.06	EOF