8.29/8.23	MAYBE
8.29/8.23	
8.29/8.23	DP problem for innermost termination.
8.29/8.23	P =
8.29/8.23	  f7#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
8.29/8.23	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f3#(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
8.29/8.23	  f2#(I14, I15, I16, I17, I18, I19, I20) -> f4#(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
8.29/8.23	  f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 
8.29/8.23	  f3#(I29, I30, I31, I32, I33, I34, I35) -> f5#(I29, I30, I31, I32, I33, I34, I34 + I35) [0 <= I35]
8.29/8.23	  f3#(I36, I37, I38, I39, I40, I41, I42) -> f4#(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
8.29/8.23	  f1#(I45, I46, I47, I48, I49, I50, I51) -> f2#(I45, I46, I47, I48, I49, I50, I51) 
8.29/8.23	R = 
8.29/8.23	  f7(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
8.29/8.23	  f4(I0, I1, I2, I3, I4, I5, I6) -> f6(rnd1, rnd2, I2, 0, I4, I5, I6) [rnd1 = rnd2 /\ rnd2 = 0]
8.29/8.23	  f2(I7, I8, I9, I10, I11, I12, I13) -> f3(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
8.29/8.23	  f2(I14, I15, I16, I17, I18, I19, I20) -> f4(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
8.29/8.23	  f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) 
8.29/8.23	  f3(I29, I30, I31, I32, I33, I34, I35) -> f5(I29, I30, I31, I32, I33, I34, I34 + I35) [0 <= I35]
8.29/8.23	  f3(I36, I37, I38, I39, I40, I41, I42) -> f4(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
8.29/8.23	  f1(I45, I46, I47, I48, I49, I50, I51) -> f2(I45, I46, I47, I48, I49, I50, I51) 
8.29/8.23	
8.29/8.23	The dependency graph for this problem is:
8.29/8.23	  0 -> 6
8.29/8.23	  1 -> 4, 5
8.29/8.23	  2 -> 
8.29/8.23	  3 -> 4, 5
8.29/8.23	  4 -> 3
8.29/8.23	  5 -> 
8.29/8.23	  6 -> 1, 2
8.29/8.23	Where:
8.29/8.23	  0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 
8.29/8.23	  1) f2#(I7, I8, I9, I10, I11, I12, I13) -> f3#(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
8.29/8.23	  2) f2#(I14, I15, I16, I17, I18, I19, I20) -> f4#(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
8.29/8.23	  3) f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 
8.29/8.23	  4) f3#(I29, I30, I31, I32, I33, I34, I35) -> f5#(I29, I30, I31, I32, I33, I34, I34 + I35) [0 <= I35]
8.29/8.23	  5) f3#(I36, I37, I38, I39, I40, I41, I42) -> f4#(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
8.29/8.23	  6) f1#(I45, I46, I47, I48, I49, I50, I51) -> f2#(I45, I46, I47, I48, I49, I50, I51) 
8.29/8.23	
8.29/8.23	We have the following SCCs.
8.29/8.23	  { 3, 4 }
8.29/8.23	
8.29/8.23	DP problem for innermost termination.
8.29/8.23	P =
8.29/8.23	  f5#(I22, I23, I24, I25, I26, I27, I28) -> f3#(I22, I23, I24, I25, I26, I27, I28) 
8.29/8.23	  f3#(I29, I30, I31, I32, I33, I34, I35) -> f5#(I29, I30, I31, I32, I33, I34, I34 + I35) [0 <= I35]
8.29/8.23	R = 
8.29/8.23	  f7(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) 
8.29/8.23	  f4(I0, I1, I2, I3, I4, I5, I6) -> f6(rnd1, rnd2, I2, 0, I4, I5, I6) [rnd1 = rnd2 /\ rnd2 = 0]
8.29/8.23	  f2(I7, I8, I9, I10, I11, I12, I13) -> f3(I7, I8, I9, I10, I11, I12, I13) [0 <= -1 - I12]
8.29/8.23	  f2(I14, I15, I16, I17, I18, I19, I20) -> f4(I21, I15, rnd3, I17, 0, I19, I20) [I21 = rnd3 /\ rnd3 = 0 /\ -1 * I19 <= 0]
8.29/8.23	  f5(I22, I23, I24, I25, I26, I27, I28) -> f3(I22, I23, I24, I25, I26, I27, I28) 
8.29/8.23	  f3(I29, I30, I31, I32, I33, I34, I35) -> f5(I29, I30, I31, I32, I33, I34, I34 + I35) [0 <= I35]
8.29/8.23	  f3(I36, I37, I38, I39, I40, I41, I42) -> f4(I43, I37, I44, I39, 0, I41, I42) [I43 = I44 /\ I44 = 0 /\ 1 + I42 <= 0]
8.29/8.23	  f1(I45, I46, I47, I48, I49, I50, I51) -> f2(I45, I46, I47, I48, I49, I50, I51) 
8.29/8.23	
8.29/11.21	EOF