0.00/0.56	YES
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  init#(x1, x2) -> f3#(rnd1, rnd2) 
0.00/0.56	  f6#(I0, I1) -> f6#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	  f3#(I2, I3) -> f6#(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.00/0.56	  f5#(I5, I6) -> f5#(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	  f4#(I8, I9) -> f4#(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	  f4#(I11, I12) -> f5#(I13, I14) [I13 + 2 <= I11 /\ I12 <= 1 /\ 0 <= I11 - 1]
0.00/0.56	  f2#(I15, I16) -> f4#(I17, I18) [-1 <= I18 - 1 /\ 0 <= y1 - 1 /\ I17 <= I15 /\ 0 <= I15 - 1 /\ 0 <= I17 - 1]
0.00/0.56	  f3#(I19, I20) -> f2#(I21, I22) [-1 <= I21 - 1 /\ 0 <= I19 - 1]
0.00/0.56	  f1#(I23, I24) -> f2#(I25, I26) [-1 <= I25 - 1 /\ -1 <= I23 - 1 /\ I25 <= I23]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.56	  f6(I0, I1) -> f6(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	  f3(I2, I3) -> f6(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.00/0.56	  f5(I5, I6) -> f5(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	  f4(I8, I9) -> f4(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	  f4(I11, I12) -> f5(I13, I14) [I13 + 2 <= I11 /\ I12 <= 1 /\ 0 <= I11 - 1]
0.00/0.56	  f2(I15, I16) -> f4(I17, I18) [-1 <= I18 - 1 /\ 0 <= y1 - 1 /\ I17 <= I15 /\ 0 <= I15 - 1 /\ 0 <= I17 - 1]
0.00/0.56	  f3(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ 0 <= I19 - 1]
0.00/0.56	  f1(I23, I24) -> f2(I25, I26) [-1 <= I25 - 1 /\ -1 <= I23 - 1 /\ I25 <= I23]
0.00/0.56	
0.00/0.56	The dependency graph for this problem is:
0.00/0.56	  0 -> 2, 7
0.00/0.56	  1 -> 1
0.00/0.56	  2 -> 1
0.00/0.56	  3 -> 3
0.00/0.56	  4 -> 4, 5
0.00/0.56	  5 -> 3
0.00/0.56	  6 -> 4, 5
0.00/0.56	  7 -> 6
0.00/0.56	  8 -> 6
0.00/0.56	Where:
0.00/0.56	  0) init#(x1, x2) -> f3#(rnd1, rnd2) 
0.00/0.56	  1) f6#(I0, I1) -> f6#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	  2) f3#(I2, I3) -> f6#(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.00/0.56	  3) f5#(I5, I6) -> f5#(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	  4) f4#(I8, I9) -> f4#(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	  5) f4#(I11, I12) -> f5#(I13, I14) [I13 + 2 <= I11 /\ I12 <= 1 /\ 0 <= I11 - 1]
0.00/0.56	  6) f2#(I15, I16) -> f4#(I17, I18) [-1 <= I18 - 1 /\ 0 <= y1 - 1 /\ I17 <= I15 /\ 0 <= I15 - 1 /\ 0 <= I17 - 1]
0.00/0.56	  7) f3#(I19, I20) -> f2#(I21, I22) [-1 <= I21 - 1 /\ 0 <= I19 - 1]
0.00/0.56	  8) f1#(I23, I24) -> f2#(I25, I26) [-1 <= I25 - 1 /\ -1 <= I23 - 1 /\ I25 <= I23]
0.00/0.56	
0.00/0.56	We have the following SCCs.
0.00/0.56	  { 1 }
0.00/0.56	  { 4 }
0.00/0.56	  { 3 }
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  f5#(I5, I6) -> f5#(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.56	  f6(I0, I1) -> f6(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	  f3(I2, I3) -> f6(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.00/0.56	  f5(I5, I6) -> f5(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	  f4(I8, I9) -> f4(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	  f4(I11, I12) -> f5(I13, I14) [I13 + 2 <= I11 /\ I12 <= 1 /\ 0 <= I11 - 1]
0.00/0.56	  f2(I15, I16) -> f4(I17, I18) [-1 <= I18 - 1 /\ 0 <= y1 - 1 /\ I17 <= I15 /\ 0 <= I15 - 1 /\ 0 <= I17 - 1]
0.00/0.56	  f3(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ 0 <= I19 - 1]
0.00/0.56	  f1(I23, I24) -> f2(I25, I26) [-1 <= I25 - 1 /\ -1 <= I23 - 1 /\ I25 <= I23]
0.00/0.56	
0.00/0.56	We use the basic value criterion with the projection function NU:
0.00/0.56	  NU[f5#(z1,z2)] = z1
0.00/0.56	
0.00/0.56	This gives the following inequalities:
0.00/0.56	  0 <= I5 - 1  ==>  I5 >! I5 - 1
0.00/0.56	
0.00/0.56	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  f4#(I8, I9) -> f4#(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.56	  f6(I0, I1) -> f6(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	  f3(I2, I3) -> f6(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.00/0.56	  f5(I5, I6) -> f5(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	  f4(I8, I9) -> f4(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	  f4(I11, I12) -> f5(I13, I14) [I13 + 2 <= I11 /\ I12 <= 1 /\ 0 <= I11 - 1]
0.00/0.56	  f2(I15, I16) -> f4(I17, I18) [-1 <= I18 - 1 /\ 0 <= y1 - 1 /\ I17 <= I15 /\ 0 <= I15 - 1 /\ 0 <= I17 - 1]
0.00/0.56	  f3(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ 0 <= I19 - 1]
0.00/0.56	  f1(I23, I24) -> f2(I25, I26) [-1 <= I25 - 1 /\ -1 <= I23 - 1 /\ I25 <= I23]
0.00/0.56	
0.00/0.56	We use the basic value criterion with the projection function NU:
0.00/0.56	  NU[f4#(z1,z2)] = z2
0.00/0.56	
0.00/0.56	This gives the following inequalities:
0.00/0.56	  -1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8  ==>  I9 >! I9 - 1
0.00/0.56	
0.00/0.56	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/0.56	
0.00/0.56	DP problem for innermost termination.
0.00/0.56	P =
0.00/0.56	  f6#(I0, I1) -> f6#(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	R = 
0.00/0.56	  init(x1, x2) -> f3(rnd1, rnd2) 
0.00/0.56	  f6(I0, I1) -> f6(I0 - 1, I1 + 1) [0 <= I1 - 1 /\ 0 <= I0 - 1]
0.00/0.56	  f3(I2, I3) -> f6(I4, 1) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1]
0.00/0.56	  f5(I5, I6) -> f5(I5 - 1, I7) [0 <= I5 - 1]
0.00/0.56	  f4(I8, I9) -> f4(I10, I9 - 1) [-1 <= I10 - 1 /\ 0 <= I8 - 1 /\ 1 <= I9 - 1 /\ I10 + 1 <= I8]
0.00/0.56	  f4(I11, I12) -> f5(I13, I14) [I13 + 2 <= I11 /\ I12 <= 1 /\ 0 <= I11 - 1]
0.00/0.56	  f2(I15, I16) -> f4(I17, I18) [-1 <= I18 - 1 /\ 0 <= y1 - 1 /\ I17 <= I15 /\ 0 <= I15 - 1 /\ 0 <= I17 - 1]
0.00/0.56	  f3(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ 0 <= I19 - 1]
0.00/0.56	  f1(I23, I24) -> f2(I25, I26) [-1 <= I25 - 1 /\ -1 <= I23 - 1 /\ I25 <= I23]
0.00/0.56	
0.00/0.56	We use the basic value criterion with the projection function NU:
0.00/0.56	  NU[f6#(z1,z2)] = z1
0.00/0.56	
0.00/0.56	This gives the following inequalities:
0.00/0.56	  0 <= I1 - 1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - 1
0.00/0.56	
0.00/0.56	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
0.00/3.54	EOF