1.35/1.65	MAYBE
1.35/1.65	
1.35/1.65	DP problem for innermost termination.
1.35/1.65	P =
1.35/1.65	  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
1.35/1.65	  f6#(I0, I1, I2, I3) -> f6#(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	  f5#(I8, I9, I10, I11) -> f5#(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	  f5#(I15, I16, I17, I18) -> f6#(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1]
1.35/1.65	  f4#(I23, I24, I25, I26) -> f4#(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	  f4#(I29, I30, I31, I32) -> f5#(2, I33, I34, I35) [I30 <= 0]
1.35/1.65	  f3#(I36, I37, I38, I39) -> f4#(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0]
1.35/1.65	  f3#(I42, I43, I44, I45) -> f3#(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42]
1.35/1.65	  f2#(I47, I48, I49, I50) -> f3#(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0]
1.35/1.65	  f2#(I52, I53, I54, I55) -> f2#(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52]
1.35/1.65	  f1#(I57, I58, I59, I60) -> f2#(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]
1.35/1.65	R = 
1.35/1.65	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.35/1.65	  f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	  f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	  f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1]
1.35/1.65	  f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	  f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0]
1.35/1.65	  f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0]
1.35/1.65	  f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42]
1.35/1.65	  f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0]
1.35/1.65	  f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52]
1.35/1.65	  f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]
1.35/1.65	
1.35/1.65	The dependency graph for this problem is:
1.35/1.65	  0 -> 10
1.35/1.65	  1 -> 1
1.35/1.65	  2 -> 2, 3
1.35/1.65	  3 -> 1
1.35/1.65	  4 -> 4, 5
1.35/1.65	  5 -> 2, 3
1.35/1.65	  6 -> 4, 5
1.35/1.65	  7 -> 6, 7
1.35/1.65	  8 -> 6, 7
1.35/1.65	  9 -> 8, 9
1.35/1.65	  10 -> 8, 9
1.35/1.65	Where:
1.35/1.65	  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
1.35/1.65	  1) f6#(I0, I1, I2, I3) -> f6#(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	  2) f5#(I8, I9, I10, I11) -> f5#(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	  3) f5#(I15, I16, I17, I18) -> f6#(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1]
1.35/1.65	  4) f4#(I23, I24, I25, I26) -> f4#(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	  5) f4#(I29, I30, I31, I32) -> f5#(2, I33, I34, I35) [I30 <= 0]
1.35/1.65	  6) f3#(I36, I37, I38, I39) -> f4#(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0]
1.35/1.65	  7) f3#(I42, I43, I44, I45) -> f3#(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42]
1.35/1.65	  8) f2#(I47, I48, I49, I50) -> f3#(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0]
1.35/1.65	  9) f2#(I52, I53, I54, I55) -> f2#(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52]
1.35/1.65	  10) f1#(I57, I58, I59, I60) -> f2#(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]
1.35/1.65	
1.35/1.65	We have the following SCCs.
1.35/1.65	  { 9 }
1.35/1.65	  { 7 }
1.35/1.65	  { 4 }
1.35/1.65	  { 2 }
1.35/1.65	  { 1 }
1.35/1.65	
1.35/1.65	DP problem for innermost termination.
1.35/1.65	P =
1.35/1.65	  f6#(I0, I1, I2, I3) -> f6#(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	R = 
1.35/1.65	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.35/1.65	  f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	  f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	  f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1]
1.35/1.65	  f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	  f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0]
1.35/1.65	  f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0]
1.35/1.65	  f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42]
1.35/1.65	  f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0]
1.35/1.65	  f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52]
1.35/1.65	  f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]
1.35/1.65	
1.35/1.65	We use the basic value criterion with the projection function NU:
1.35/1.65	  NU[f6#(z1,z2,z3,z4)] = z1
1.35/1.65	
1.35/1.65	This gives the following inequalities:
1.35/1.65	  -1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0  ==>  I0 >! I4
1.35/1.65	
1.35/1.65	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.35/1.65	
1.35/1.65	DP problem for innermost termination.
1.35/1.65	P =
1.35/1.65	  f5#(I8, I9, I10, I11) -> f5#(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	R = 
1.35/1.65	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.35/1.65	  f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	  f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	  f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1]
1.35/1.65	  f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	  f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0]
1.35/1.65	  f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0]
1.35/1.65	  f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42]
1.35/1.65	  f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0]
1.35/1.65	  f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52]
1.35/1.65	  f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]
1.35/1.65	
1.35/1.65	We use the basic value criterion with the projection function NU:
1.35/1.65	  NU[f5#(z1,z2,z3,z4)] = z1
1.35/1.65	
1.35/1.65	This gives the following inequalities:
1.35/1.65	  I8 <= 2 /\ -1 <= I8 - 1  ==>  I8 >! I8 - 1
1.35/1.65	
1.35/1.65	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
1.35/1.65	
1.35/1.65	DP problem for innermost termination.
1.35/1.65	P =
1.35/1.65	  f4#(I23, I24, I25, I26) -> f4#(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	R = 
1.35/1.65	  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
1.35/1.65	  f6(I0, I1, I2, I3) -> f6(I4, I5, I6, I7) [-1 <= I4 - 1 /\ 0 <= I0 - 1 /\ I4 + 1 <= I0]
1.35/1.65	  f5(I8, I9, I10, I11) -> f5(I8 - 1, I12, I13, I14) [I8 <= 2 /\ -1 <= I8 - 1]
1.35/1.65	  f5(I15, I16, I17, I18) -> f6(I19, I20, I21, I22) [I15 <= 2 /\ -1 <= I19 - 1 /\ -1 <= I15 - 1]
1.35/1.65	  f4(I23, I24, I25, I26) -> f4(I23 - 1, I23, I27, I28) [0 <= I24 - 1]
1.35/1.65	  f4(I29, I30, I31, I32) -> f5(2, I33, I34, I35) [I30 <= 0]
1.35/1.65	  f3(I36, I37, I38, I39) -> f4(I39 - 1, I39, I40, I41) [0 <= I36 - 1 /\ -1 <= I39 - 1 /\ I38 <= 0]
1.35/1.65	  f3(I42, I43, I44, I45) -> f3(I46, I43 - 1, I43, I45) [0 <= I46 - 1 /\ 0 <= I42 - 1 /\ 0 <= I44 - 1 /\ I46 <= I42]
1.35/1.65	  f2(I47, I48, I49, I50) -> f3(I51, I50 - 1, I50, I50) [0 <= I51 - 1 /\ 0 <= I47 - 1 /\ I51 <= I47 /\ -1 <= I50 - 1 /\ I49 <= 0]
1.35/1.65	  f2(I52, I53, I54, I55) -> f2(I56, I53 - 1, I53, I55) [0 <= I56 - 1 /\ 0 <= I52 - 1 /\ 0 <= I54 - 1 /\ I56 <= I52]
1.35/1.65	  f1(I57, I58, I59, I60) -> f2(I61, I58 - 1, I58, I58) [0 <= I61 - 1 /\ 0 <= I57 - 1 /\ -1 <= I58 - 1 /\ I61 <= I57]
1.35/1.65	
1.35/1.65	EOF