1.53/1.58	MAYBE
1.53/1.58	
1.53/1.58	DP problem for innermost termination.
1.53/1.58	P =
1.53/1.58	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
1.53/1.58	  f3#(I0, I1, I2, I3) -> f3#(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2]
1.53/1.58	  f3#(I7, I8, I9, I10) -> f3#(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1]
1.53/1.58	  f2#(I15, I16, I17, I18) -> f3#(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0]
1.53/1.58	  f2#(I19, I20, I21, I22) -> f2#(I19 - 1, I19, 1, I23) [0 <= I20 - 1]
1.53/1.58	  f2#(I24, I25, I26, I27) -> f2#(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1]
1.53/1.58	  f1#(I30, I31, I32, I33) -> f2#(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1]
1.53/1.58	R = 
1.53/1.58	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.53/1.58	  f3(I0, I1, I2, I3) -> f3(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2]
1.53/1.58	  f3(I7, I8, I9, I10) -> f3(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1]
1.53/1.58	  f2(I15, I16, I17, I18) -> f3(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0]
1.53/1.58	  f2(I19, I20, I21, I22) -> f2(I19 - 1, I19, 1, I23) [0 <= I20 - 1]
1.53/1.58	  f2(I24, I25, I26, I27) -> f2(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1]
1.53/1.58	  f1(I30, I31, I32, I33) -> f2(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1]
1.53/1.58	
1.53/1.58	The dependency graph for this problem is:
1.53/1.58	  0 -> 6
1.53/1.58	  1 -> 1, 2
1.53/1.58	  2 -> 1, 2
1.53/1.58	  3 -> 1
1.53/1.58	  4 -> 3, 4, 5
1.53/1.58	  5 -> 3, 4, 5
1.53/1.58	  6 -> 4
1.53/1.58	Where:
1.53/1.58	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
1.53/1.58	  1) f3#(I0, I1, I2, I3) -> f3#(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2]
1.53/1.58	  2) f3#(I7, I8, I9, I10) -> f3#(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1]
1.53/1.58	  3) f2#(I15, I16, I17, I18) -> f3#(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0]
1.53/1.58	  4) f2#(I19, I20, I21, I22) -> f2#(I19 - 1, I19, 1, I23) [0 <= I20 - 1]
1.53/1.58	  5) f2#(I24, I25, I26, I27) -> f2#(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1]
1.53/1.58	  6) f1#(I30, I31, I32, I33) -> f2#(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1]
1.53/1.58	
1.53/1.58	We have the following SCCs.
1.53/1.58	  { 4, 5 }
1.53/1.58	  { 1, 2 }
1.53/1.58	
1.53/1.58	DP problem for innermost termination.
1.53/1.58	P =
1.53/1.58	  f3#(I0, I1, I2, I3) -> f3#(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2]
1.53/1.58	  f3#(I7, I8, I9, I10) -> f3#(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1]
1.53/1.58	R = 
1.53/1.58	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.53/1.58	  f3(I0, I1, I2, I3) -> f3(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2]
1.53/1.58	  f3(I7, I8, I9, I10) -> f3(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1]
1.53/1.58	  f2(I15, I16, I17, I18) -> f3(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0]
1.53/1.58	  f2(I19, I20, I21, I22) -> f2(I19 - 1, I19, 1, I23) [0 <= I20 - 1]
1.53/1.58	  f2(I24, I25, I26, I27) -> f2(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1]
1.53/1.58	  f1(I30, I31, I32, I33) -> f2(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1]
1.53/1.58	
1.53/1.58	We use the basic value criterion with the projection function NU:
1.53/1.58	  NU[f3#(z1,z2,z3,z4)] = z3
1.53/1.58	
1.53/1.58	This gives the following inequalities:
1.53/1.58	  I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2  ==>  I2 >! I5
1.53/1.58	  0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1  ==>  I9 >! I12
1.53/1.58	
1.53/1.58	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.53/1.58	
1.53/1.58	DP problem for innermost termination.
1.53/1.58	P =
1.53/1.58	  f2#(I19, I20, I21, I22) -> f2#(I19 - 1, I19, 1, I23) [0 <= I20 - 1]
1.53/1.58	  f2#(I24, I25, I26, I27) -> f2#(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1]
1.53/1.58	R = 
1.53/1.58	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.53/1.58	  f3(I0, I1, I2, I3) -> f3(1, I4, I5, I6) [I5 <= I1 - 1 /\ 0 <= I1 - 1 /\ I6 <= y1 - 1 /\ -1 <= y1 - 1 /\ I1 = I2]
1.53/1.59	  f3(I7, I8, I9, I10) -> f3(I11, I8, I12, I13) [0 <= I7 - 1 /\ 0 <= I10 - 1 /\ 0 <= I9 - 1 /\ I7 <= I11 - 1 /\ I7 <= I8 - 1 /\ I12 <= I9 - 1 /\ I13 <= I10 - 1 /\ I7 <= I14 - 1]
1.53/1.59	  f2(I15, I16, I17, I18) -> f3(0, I17, I17, 0) [0 <= I17 - 1 /\ I16 <= 0]
1.53/1.59	  f2(I19, I20, I21, I22) -> f2(I19 - 1, I19, 1, I23) [0 <= I20 - 1]
1.53/1.59	  f2(I24, I25, I26, I27) -> f2(I24 - 1, I24, I28, I29) [0 <= I26 - 1 /\ I26 <= I28 - 1 /\ 0 <= I25 - 1]
1.53/1.59	  f1(I30, I31, I32, I33) -> f2(I31 - 1, I31, 0, I34) [-1 <= I31 - 1 /\ 0 <= I30 - 1]
1.53/1.59	
1.53/4.56	EOF