13.67/13.48	YES
13.67/13.48	
13.67/13.48	DP problem for innermost termination.
13.67/13.48	P =
13.67/13.48	  init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
13.67/13.48	  f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  f3#(I31, I32, I33, I34, I35) -> f2#(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
13.67/13.48	  f2#(I38, I39, I40, I41, I42) -> f3#(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	  f1#(I43, I44, I45, I46, I47) -> f2#(0, 0, 0, I48, I49) 
13.67/13.48	R = 
13.67/13.48	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
13.67/13.48	  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
13.67/13.48	  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 
13.67/13.48	
13.67/13.48	The dependency graph for this problem is:
13.67/13.48	  0 -> 8
13.67/13.48	  1 -> 5
13.67/13.48	  2 -> 7
13.67/13.48	  3 -> 1, 2, 3, 4
13.67/13.48	  4 -> 1, 2, 3, 4
13.67/13.48	  5 -> 7
13.67/13.48	  6 -> 7
13.67/13.48	  7 -> 1, 3, 4
13.67/13.48	  8 -> 7
13.67/13.48	Where:
13.67/13.48	  0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 
13.67/13.48	  1) f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  2) f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  3) f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  4) f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  5) f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  6) f3#(I31, I32, I33, I34, I35) -> f2#(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
13.67/13.48	  7) f2#(I38, I39, I40, I41, I42) -> f3#(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	  8) f1#(I43, I44, I45, I46, I47) -> f2#(0, 0, 0, I48, I49) 
13.67/13.48	
13.67/13.48	We have the following SCCs.
13.67/13.48	  { 1, 2, 3, 4, 5, 7 }
13.67/13.48	
13.67/13.48	DP problem for innermost termination.
13.67/13.48	P =
13.67/13.48	  f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  f2#(I38, I39, I40, I41, I42) -> f3#(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	R = 
13.67/13.48	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
13.67/13.48	  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
13.67/13.48	  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 
13.67/13.48	
13.67/13.48	We use the reverse value criterion with the projection function NU:
13.67/13.48	  NU[f2#(z1,z2,z3,z4,z5)] = 19 + -1 * z2
13.67/13.48	  NU[f3#(z1,z2,z3,z4,z5)] = 19 + -1 * (z2 + 1)
13.67/13.48	
13.67/13.48	This gives the following inequalities:
13.67/13.48	  I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19  ==>  19 + -1 * (I1 + 1) >= 19 + -1 * (I1 + 1)
13.67/13.48	  I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19  ==>  19 + -1 * (I6 + 1) >= 19 + -1 * (I6 + 1)
13.67/13.48	  I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16  ==>  19 + -1 * (I13 + 1) >= 19 + -1 * (I13 + 1)
13.67/13.48	  I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21  ==>  19 + -1 * (I18 + 1) >= 19 + -1 * (I18 + 1)
13.67/13.48	  I27 = I28 /\ 1 = I26 /\ I27 <= 19  ==>  19 + -1 * (I25 + 1) >= 19 + -1 * (I25 + 1)
13.67/13.48	  I39 = I40 /\ I39 <= 19  ==>  19 + -1 * I39 > 19 + -1 * (I39 + 1)  with  19 + -1 * I39 >= 0
13.67/13.48	
13.67/13.48	We remove all the strictly oriented dependency pairs.
13.67/13.48	
13.67/13.48	DP problem for innermost termination.
13.67/13.48	P =
13.67/13.48	  f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	R = 
13.67/13.48	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
13.67/13.48	  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
13.67/13.48	  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 
13.67/13.48	
13.67/13.48	The dependency graph for this problem is:
13.67/13.48	  1 -> 5
13.67/13.48	  2 -> 
13.67/13.48	  3 -> 1, 2, 3, 4
13.67/13.48	  4 -> 1, 2, 3, 4
13.67/13.48	  5 -> 
13.67/13.48	Where:
13.67/13.48	  1) f3#(I0, I1, I2, I3, I4) -> f3#(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  2) f3#(I5, I6, I7, I8, I9) -> f2#(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  3) f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  4) f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  5) f3#(I24, I25, I26, I27, I28) -> f2#(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	
13.67/13.48	We have the following SCCs.
13.67/13.48	  { 3, 4 }
13.67/13.48	
13.67/13.48	DP problem for innermost termination.
13.67/13.48	P =
13.67/13.48	  f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	R = 
13.67/13.48	  init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) 
13.67/13.48	  f3(I0, I1, I2, I3, I4) -> f3(I0, I1, 1, I3, I3) [I3 = I4 /\ 0 = I2 /\ I1 <= 19 /\ I3 <= 19]
13.67/13.48	  f3(I5, I6, I7, I8, I9) -> f2(I5 + 1, I6 + 1, I6 + 1, I10, I11) [I8 = I9 /\ 0 = I7 /\ 19 <= I8 - 1 /\ I6 <= 19 /\ I5 <= 19]
13.67/13.48	  f3(I12, I13, I14, I15, I16) -> f3(I12, I13, 0, I15 + 1, I15 + 1) [I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16]
13.67/13.48	  f3(I17, I18, I19, I20, I21) -> f3(I17, I18, 0, I20 + 1, I20 + 1) [I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21]
13.67/13.48	  f3(I24, I25, I26, I27, I28) -> f2(I24, I25 + 1, I25 + 1, I29, I30) [I27 = I28 /\ 1 = I26 /\ I27 <= 19]
13.67/13.48	  f3(I31, I32, I33, I34, I35) -> f2(I31, I32 + 1, I32 + 1, I36, I37) [I34 = I35 /\ 1 = I33 /\ 19 <= I34 - 1]
13.67/13.48	  f2(I38, I39, I40, I41, I42) -> f3(I38, I39, 0, 0, 0) [I39 = I40 /\ I39 <= 19]
13.67/13.48	  f1(I43, I44, I45, I46, I47) -> f2(0, 0, 0, I48, I49) 
13.67/13.48	
13.67/13.48	We use the reverse value criterion with the projection function NU:
13.67/13.48	  NU[f3#(z1,z2,z3,z4,z5)] = 19 + -1 * z4
13.67/13.48	
13.67/13.48	This gives the following inequalities:
13.67/13.48	  I13 <= 19 /\ y1 <= y2 - 1 /\ I15 <= 19 /\ 0 = I14 /\ I15 = I16  ==>  19 + -1 * I15 > 19 + -1 * (I15 + 1)  with  19 + -1 * I15 >= 0
13.67/13.48	  I18 <= 19 /\ I22 <= I23 - 1 /\ I20 <= 19 /\ 0 = I19 /\ I20 = I21  ==>  19 + -1 * I20 > 19 + -1 * (I20 + 1)  with  19 + -1 * I20 >= 0
13.67/13.48	
13.67/13.48	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
13.67/13.48	EOF