4.65/5.02	MAYBE
4.65/5.02	
4.65/5.02	DP problem for innermost termination.
4.65/5.02	P =
4.65/5.02	  f10#(x1, x2, x3) -> f1#(x1, x2, x3) 
4.65/5.02	  f9#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.65/5.02	  f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f5#(I18, I19, I20) -> f6#(I18, I19, I20) [1 <= I19]
4.65/5.02	  f5#(I21, I22, I23) -> f6#(I21, I22, I23) [1 + I22 <= 0]
4.65/5.02	  f2#(I24, I25, I26) -> f5#(I24, I27, I26) [I27 = I27]
4.65/5.02	  f4#(I28, I29, I30) -> f2#(I28, I29, I30) 
4.65/5.02	  f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	  f1#(I40, I41, I42) -> f2#(I40, I41, I42) 
4.65/5.02	R = 
4.65/5.02	  f10(x1, x2, x3) -> f1(x1, x2, x3) 
4.65/5.02	  f9(I0, I1, I2) -> f2(I0, I1, I2) 
4.65/5.02	  f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0]
4.65/5.02	  f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19]
4.65/5.02	  f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0]
4.65/5.02	  f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27]
4.65/5.02	  f4(I28, I29, I30) -> f2(I28, I29, I30) 
4.65/5.02	  f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	  f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39]
4.65/5.02	  f1(I40, I41, I42) -> f2(I40, I41, I42) 
4.65/5.02	
4.65/5.02	The dependency graph for this problem is:
4.65/5.02	  0 -> 11
4.65/5.02	  1 -> 5, 8, 10
4.65/5.02	  2 -> 1
4.65/5.02	  3 -> 2
4.65/5.02	  4 -> 2
4.65/5.02	  5 -> 3, 4
4.65/5.02	  6 -> 
4.65/5.02	  7 -> 
4.65/5.02	  8 -> 6, 7
4.65/5.02	  9 -> 5, 8, 10
4.65/5.02	  10 -> 9
4.65/5.02	  11 -> 5, 8, 10
4.65/5.02	Where:
4.65/5.02	  0) f10#(x1, x2, x3) -> f1#(x1, x2, x3) 
4.65/5.02	  1) f9#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.65/5.02	  2) f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  3) f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7]
4.65/5.02	  4) f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  5) f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  6) f5#(I18, I19, I20) -> f6#(I18, I19, I20) [1 <= I19]
4.65/5.02	  7) f5#(I21, I22, I23) -> f6#(I21, I22, I23) [1 + I22 <= 0]
4.65/5.02	  8) f2#(I24, I25, I26) -> f5#(I24, I27, I26) [I27 = I27]
4.65/5.02	  9) f4#(I28, I29, I30) -> f2#(I28, I29, I30) 
4.65/5.02	  10) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	  11) f1#(I40, I41, I42) -> f2#(I40, I41, I42) 
4.65/5.02	
4.65/5.02	We have the following SCCs.
4.65/5.02	  { 1, 2, 3, 4, 5, 9, 10 }
4.65/5.02	
4.65/5.02	DP problem for innermost termination.
4.65/5.02	P =
4.65/5.02	  f9#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.65/5.02	  f8#(I3, I4, I5) -> f9#(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f4#(I28, I29, I30) -> f2#(I28, I29, I30) 
4.65/5.02	  f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	R = 
4.65/5.02	  f10(x1, x2, x3) -> f1(x1, x2, x3) 
4.65/5.02	  f9(I0, I1, I2) -> f2(I0, I1, I2) 
4.65/5.02	  f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0]
4.65/5.02	  f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19]
4.65/5.02	  f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0]
4.65/5.02	  f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27]
4.65/5.02	  f4(I28, I29, I30) -> f2(I28, I29, I30) 
4.65/5.02	  f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	  f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39]
4.65/5.02	  f1(I40, I41, I42) -> f2(I40, I41, I42) 
4.65/5.02	
4.65/5.02	We use the basic value criterion with the projection function NU:
4.65/5.02	  NU[f4#(z1,z2,z3)] = z3
4.65/5.02	  NU[f7#(z1,z2,z3)] = z3
4.65/5.02	  NU[f8#(z1,z2,z3)] = z3
4.65/5.02	  NU[f2#(z1,z2,z3)] = z3
4.65/5.02	  NU[f9#(z1,z2,z3)] = z3
4.65/5.02	
4.65/5.02	This gives the following inequalities:
4.65/5.02	    ==>  I2 (>! \union =) I2
4.65/5.02	  0 <= -1 + -1 + I5  ==>  I5 >! -1 + I5
4.65/5.02	  1 <= I7  ==>  I8 (>! \union =) I8
4.65/5.02	  1 + I10 <= 0  ==>  I11 (>! \union =) I11
4.65/5.02	  rnd2 = rnd2  ==>  I14 (>! \union =) I14
4.65/5.02	    ==>  I30 (>! \union =) I30
4.65/5.02	  0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34  ==>  I33 (>! \union =) I33
4.65/5.02	
4.65/5.02	We remove all the strictly oriented dependency pairs.
4.65/5.02	
4.65/5.02	DP problem for innermost termination.
4.65/5.02	P =
4.65/5.02	  f9#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.65/5.02	  f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f4#(I28, I29, I30) -> f2#(I28, I29, I30) 
4.65/5.02	  f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	R = 
4.65/5.02	  f10(x1, x2, x3) -> f1(x1, x2, x3) 
4.65/5.02	  f9(I0, I1, I2) -> f2(I0, I1, I2) 
4.65/5.02	  f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0]
4.65/5.02	  f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19]
4.65/5.02	  f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0]
4.65/5.02	  f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27]
4.65/5.02	  f4(I28, I29, I30) -> f2(I28, I29, I30) 
4.65/5.02	  f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	  f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39]
4.65/5.02	  f1(I40, I41, I42) -> f2(I40, I41, I42) 
4.65/5.02	
4.65/5.02	The dependency graph for this problem is:
4.65/5.02	  1 -> 5, 10
4.65/5.02	  3 -> 
4.65/5.02	  4 -> 
4.65/5.02	  5 -> 3, 4
4.65/5.02	  9 -> 5, 10
4.65/5.02	  10 -> 9
4.65/5.02	Where:
4.65/5.02	  1) f9#(I0, I1, I2) -> f2#(I0, I1, I2) 
4.65/5.02	  3) f7#(I6, I7, I8) -> f8#(I6, I7, I8) [1 <= I7]
4.65/5.02	  4) f7#(I9, I10, I11) -> f8#(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  5) f2#(I12, I13, I14) -> f7#(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  9) f4#(I28, I29, I30) -> f2#(I28, I29, I30) 
4.65/5.02	  10) f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	
4.65/5.02	We have the following SCCs.
4.65/5.02	  { 9, 10 }
4.65/5.02	
4.65/5.02	DP problem for innermost termination.
4.65/5.02	P =
4.65/5.02	  f4#(I28, I29, I30) -> f2#(I28, I29, I30) 
4.65/5.02	  f2#(I31, I32, I33) -> f4#(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	R = 
4.65/5.02	  f10(x1, x2, x3) -> f1(x1, x2, x3) 
4.65/5.02	  f9(I0, I1, I2) -> f2(I0, I1, I2) 
4.65/5.02	  f8(I3, I4, I5) -> f9(I3, I4, -1 + I5) [0 <= -1 + -1 + I5]
4.65/5.02	  f7(I6, I7, I8) -> f8(I6, I7, I8) [1 <= I7]
4.65/5.02	  f7(I9, I10, I11) -> f8(I9, I10, I11) [1 + I10 <= 0]
4.65/5.02	  f2(I12, I13, I14) -> f7(I12, rnd2, I14) [rnd2 = rnd2]
4.65/5.02	  f6(I15, I16, I17) -> f3(rnd1, I16, -1 + I17) [rnd1 = rnd1 /\ -1 + I17 <= 0]
4.65/5.02	  f5(I18, I19, I20) -> f6(I18, I19, I20) [1 <= I19]
4.65/5.02	  f5(I21, I22, I23) -> f6(I21, I22, I23) [1 + I22 <= 0]
4.65/5.02	  f2(I24, I25, I26) -> f5(I24, I27, I26) [I27 = I27]
4.65/5.02	  f4(I28, I29, I30) -> f2(I28, I29, I30) 
4.65/5.02	  f2(I31, I32, I33) -> f4(I31, I34, I33) [0 <= -1 + I33 /\ 0 <= I34 /\ I34 <= 0 /\ I34 = I34]
4.65/5.02	  f2(I35, I36, I37) -> f3(I38, I39, I37) [I38 = I38 /\ I37 <= 0 /\ 0 <= I39 /\ I39 <= 0 /\ I39 = I39]
4.65/5.02	  f1(I40, I41, I42) -> f2(I40, I41, I42) 
4.65/5.02	
4.65/7.99	EOF