14.40/14.51	MAYBE
14.40/14.51	
14.40/14.51	DP problem for innermost termination.
14.40/14.51	P =
14.40/14.51	  f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
14.40/14.51	  f6#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
14.40/14.51	  f5#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
14.40/14.51	  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 <= I9]
14.40/14.51	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= 0]
14.40/14.51	  f1#(I16, I17, I18, I19) -> f4#(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.40/14.51	  f3#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) 
14.40/14.51	  f1#(I24, I25, I26, I27) -> f3#(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.40/14.51	R = 
14.40/14.51	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
14.40/14.51	  f6(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
14.40/14.51	  f5(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
14.40/14.51	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 <= I9]
14.40/14.51	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= 0]
14.40/14.51	  f1(I16, I17, I18, I19) -> f4(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.40/14.51	  f3(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) 
14.40/14.51	  f1(I24, I25, I26, I27) -> f3(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.48/14.51	  f1(I29, I30, I31, I32) -> f2(rnd1, I30, I31, I32) [rnd1 = rnd1 /\ -1 * I31 + I32 <= 0]
14.48/14.51	
14.48/14.51	The dependency graph for this problem is:
14.48/14.51	  0 -> 1
14.48/14.51	  1 -> 5, 7
14.48/14.51	  2 -> 5, 7
14.48/14.51	  3 -> 2
14.48/14.51	  4 -> 2
14.48/14.51	  5 -> 3, 4
14.48/14.51	  6 -> 5, 7
14.48/14.51	  7 -> 6
14.48/14.51	Where:
14.48/14.51	  0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
14.48/14.51	  1) f6#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
14.48/14.51	  2) f5#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
14.48/14.51	  3) f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  5) f1#(I16, I17, I18, I19) -> f4#(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  6) f3#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) 
14.48/14.51	  7) f1#(I24, I25, I26, I27) -> f3#(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.48/14.51	
14.48/14.51	We have the following SCCs.
14.48/14.51	  { 2, 3, 4, 5, 6, 7 }
14.48/14.51	
14.48/14.51	DP problem for innermost termination.
14.48/14.51	P =
14.48/14.51	  f5#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
14.48/14.51	  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  f1#(I16, I17, I18, I19) -> f4#(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  f3#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) 
14.48/14.51	  f1#(I24, I25, I26, I27) -> f3#(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.48/14.51	R = 
14.48/14.51	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
14.48/14.51	  f6(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
14.48/14.51	  f5(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
14.48/14.51	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  f1(I16, I17, I18, I19) -> f4(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  f3(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) 
14.48/14.51	  f1(I24, I25, I26, I27) -> f3(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.48/14.51	  f1(I29, I30, I31, I32) -> f2(rnd1, I30, I31, I32) [rnd1 = rnd1 /\ -1 * I31 + I32 <= 0]
14.48/14.51	
14.48/14.51	We use the extended value criterion with the projection function NU:
14.48/14.51	  NU[f3#(x0,x1,x2,x3)] = -x2 + x3 - 1
14.48/14.51	  NU[f4#(x0,x1,x2,x3)] = -x2 + x3 - 1
14.48/14.51	  NU[f1#(x0,x1,x2,x3)] = -x2 + x3 - 1
14.48/14.51	  NU[f5#(x0,x1,x2,x3)] = -x2 + x3 - 1
14.48/14.51	
14.48/14.51	This gives the following inequalities:
14.48/14.51	    ==>  -I6 + I7 - 1 >= -I6 + I7 - 1
14.48/14.51	  1 <= I9  ==>  -I10 + I11 - 1 >= -I10 + I11 - 1
14.48/14.51	  1 + I13 <= 0  ==>  -I14 + I15 - 1 >= -I14 + I15 - 1
14.48/14.51	  rnd2 = rnd2 /\ 0 <= -1 - I18 + I19  ==>  -I18 + I19 - 1 >= -I18 + I19 - 1
14.48/14.51	    ==>  -I22 + I23 - 1 >= -I22 + I23 - 1
14.48/14.51	  0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27  ==>  -I26 + I27 - 1 > -I26 + (-1 + I27) - 1  with  -I26 + I27 - 1 >= 0
14.48/14.51	
14.48/14.51	We remove all the strictly oriented dependency pairs.
14.48/14.51	
14.48/14.51	DP problem for innermost termination.
14.48/14.51	P =
14.48/14.51	  f5#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
14.48/14.51	  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  f1#(I16, I17, I18, I19) -> f4#(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  f3#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) 
14.48/14.51	R = 
14.48/14.51	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
14.48/14.51	  f6(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
14.48/14.51	  f5(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
14.48/14.51	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  f1(I16, I17, I18, I19) -> f4(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  f3(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) 
14.48/14.51	  f1(I24, I25, I26, I27) -> f3(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.48/14.51	  f1(I29, I30, I31, I32) -> f2(rnd1, I30, I31, I32) [rnd1 = rnd1 /\ -1 * I31 + I32 <= 0]
14.48/14.51	
14.48/14.51	The dependency graph for this problem is:
14.48/14.51	  2 -> 5
14.48/14.51	  3 -> 2
14.48/14.51	  4 -> 2
14.48/14.51	  5 -> 3, 4
14.48/14.51	  6 -> 5
14.48/14.51	Where:
14.48/14.51	  2) f5#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
14.48/14.51	  3) f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  5) f1#(I16, I17, I18, I19) -> f4#(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  6) f3#(I20, I21, I22, I23) -> f1#(I20, I21, I22, I23) 
14.48/14.51	
14.48/14.51	We have the following SCCs.
14.48/14.51	  { 2, 3, 4, 5 }
14.48/14.51	
14.48/14.51	DP problem for innermost termination.
14.48/14.51	P =
14.48/14.51	  f5#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
14.48/14.51	  f4#(I8, I9, I10, I11) -> f5#(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  f1#(I16, I17, I18, I19) -> f4#(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	R = 
14.48/14.51	  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
14.48/14.51	  f6(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
14.48/14.51	  f5(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
14.48/14.51	  f4(I8, I9, I10, I11) -> f5(I8, I9, I10, I11) [1 <= I9]
14.48/14.51	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 + I13 <= 0]
14.48/14.51	  f1(I16, I17, I18, I19) -> f4(I16, rnd2, I18, I19) [rnd2 = rnd2 /\ 0 <= -1 - I18 + I19]
14.48/14.51	  f3(I20, I21, I22, I23) -> f1(I20, I21, I22, I23) 
14.48/14.51	  f1(I24, I25, I26, I27) -> f3(I24, I28, I26, -1 + I27) [0 <= I28 /\ I28 <= 0 /\ I28 = I28 /\ 0 <= -1 - I26 + I27]
14.48/14.51	  f1(I29, I30, I31, I32) -> f2(rnd1, I30, I31, I32) [rnd1 = rnd1 /\ -1 * I31 + I32 <= 0]
14.48/14.51	
14.48/17.49	EOF