12.60/12.47	YES
12.60/12.47	
12.60/12.47	DP problem for innermost termination.
12.60/12.47	P =
12.60/12.47	  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
12.60/12.47	  f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
12.60/12.47	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
12.60/12.47	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	R = 
12.60/12.47	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
12.60/12.47	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
12.60/12.47	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
12.60/12.47	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
12.60/12.47	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
12.60/12.47	
12.60/12.47	The dependency graph for this problem is:
12.60/12.47	  0 -> 1
12.60/12.47	  1 -> 6, 8
12.60/12.47	  2 -> 6, 8
12.60/12.47	  3 -> 2
12.60/12.47	  4 -> 3
12.60/12.47	  5 -> 3
12.60/12.47	  6 -> 4, 5
12.60/12.47	  7 -> 6, 8
12.60/12.47	  8 -> 7
12.60/12.47	Where:
12.60/12.47	  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
12.60/12.47	  1) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 
12.60/12.47	  2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
12.60/12.47	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
12.60/12.47	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  6) f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	  8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	
12.60/12.47	We have the following SCCs.
12.60/12.47	  { 2, 3, 4, 5, 6, 7, 8 }
12.60/12.47	
12.60/12.47	DP problem for innermost termination.
12.60/12.47	P =
12.60/12.47	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
12.60/12.47	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	R = 
12.60/12.47	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
12.60/12.47	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
12.60/12.47	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
12.60/12.47	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
12.60/12.47	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
12.60/12.47	
12.60/12.47	We use the extended value criterion with the projection function NU:
12.60/12.47	  NU[f3#(x0,x1,x2,x3)] = -x2 + x3 - 1
12.60/12.47	  NU[f4#(x0,x1,x2,x3)] = -x2 + x3 - 2
12.60/12.47	  NU[f5#(x0,x1,x2,x3)] = -x2 + x3 - 2
12.60/12.47	  NU[f1#(x0,x1,x2,x3)] = -x2 + x3 - 1
12.60/12.47	  NU[f6#(x0,x1,x2,x3)] = -x2 + x3 - 1
12.60/12.47	
12.60/12.47	This gives the following inequalities:
12.60/12.47	    ==>  -I6 + I7 - 1 >= -I6 + I7 - 1
12.60/12.47	    ==>  -I10 + I11 - 2 >= -(1 + I10) + I11 - 1
12.60/12.47	  1 <= I13  ==>  -I14 + I15 - 2 >= -I14 + I15 - 2
12.60/12.47	  1 + I17 <= 0  ==>  -I18 + I19 - 2 >= -I18 + I19 - 2
12.60/12.47	  rnd2 = rnd2 /\ 0 <= -1 - I22 + I23  ==>  -I22 + I23 - 1 > -I22 + I23 - 2  with  -I22 + I23 - 1 >= 0
12.60/12.47	    ==>  -I26 + I27 - 1 >= -I26 + I27 - 1
12.60/12.47	  0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31  ==>  -I30 + I31 - 1 >= -I30 + (-1 + I31) - 1
12.60/12.47	
12.60/12.47	We remove all the strictly oriented dependency pairs.
12.60/12.47	
12.60/12.47	DP problem for innermost termination.
12.60/12.47	P =
12.60/12.47	  f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
12.60/12.47	  f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	R = 
12.60/12.47	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
12.60/12.47	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
12.60/12.47	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
12.60/12.47	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
12.60/12.47	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
12.60/12.47	
12.60/12.47	The dependency graph for this problem is:
12.60/12.47	  2 -> 8
12.60/12.47	  3 -> 2
12.60/12.47	  4 -> 3
12.60/12.47	  5 -> 3
12.60/12.47	  7 -> 8
12.60/12.47	  8 -> 7
12.60/12.47	Where:
12.60/12.47	  2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 
12.60/12.47	  3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 
12.60/12.47	  4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	  8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	
12.60/12.47	We have the following SCCs.
12.60/12.47	  { 7, 8 }
12.60/12.47	
12.60/12.47	DP problem for innermost termination.
12.60/12.47	P =
12.60/12.47	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	  f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	R = 
12.60/12.47	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
12.60/12.47	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
12.60/12.47	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
12.60/12.47	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
12.60/12.47	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
12.60/12.47	
12.60/12.47	We use the reverse value criterion with the projection function NU:
12.60/12.47	  NU[f1#(z1,z2,z3,z4)] = -1 - z3 + z4 + -1 * 0
12.60/12.47	  NU[f3#(z1,z2,z3,z4)] = -1 - z3 + z4 + -1 * 0
12.60/12.47	
12.60/12.47	This gives the following inequalities:
12.60/12.47	    ==>  -1 - I26 + I27 + -1 * 0 >= -1 - I26 + I27 + -1 * 0
12.60/12.47	  0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31  ==>  -1 - I30 + I31 + -1 * 0 > -1 - I30 + (-1 + I31) + -1 * 0  with  -1 - I30 + I31 + -1 * 0 >= 0
12.60/12.47	
12.60/12.47	We remove all the strictly oriented dependency pairs.
12.60/12.47	
12.60/12.47	DP problem for innermost termination.
12.60/12.47	P =
12.60/12.47	  f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	R = 
12.60/12.47	  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
12.60/12.47	  f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) 
12.60/12.47	  f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) 
12.60/12.47	  f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) 
12.60/12.47	  f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13]
12.60/12.47	  f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0]
12.60/12.47	  f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23]
12.60/12.47	  f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
12.60/12.47	  f1(I28, I29, I30, I31) -> f3(I28, I32, I30, -1 + I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31]
12.60/12.47	  f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0]
12.60/12.47	
12.60/12.47	The dependency graph for this problem is:
12.60/12.47	  7 -> 
12.60/12.47	Where:
12.60/12.47	  7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
12.60/12.47	
12.60/12.47	We have the following SCCs.
12.60/12.47	  
12.60/15.45	EOF