11.02/10.90	MAYBE
11.02/10.90	
11.02/10.90	DP problem for innermost termination.
11.02/10.90	P =
11.02/10.90	  f10#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
11.02/10.90	  f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
11.02/10.90	  f8#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f5#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) [I26 = I26]
11.02/10.90	  f2#(I30, I31, I32, I33, I34) -> f5#(I30, I31, I32, I35, I34) [I35 = I35]
11.02/10.90	  f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 
11.02/10.90	  f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	  f1#(I54, I55, I56, I57, I58) -> f2#(I54, I55, I56, I57, I58) 
11.02/10.90	R = 
11.02/10.90	  f10(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
11.02/10.90	  f9(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
11.02/10.90	  f8(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  f7(I10, I11, I12, I13, I14) -> f8(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f6(I20, I21, I22, I23, I24) -> f3(rnd1, I21, I22, I23, -1 + I24) [rnd1 = rnd1 /\ -1 + I24 <= 0]
11.02/10.90	  f5(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) [I26 = I26]
11.02/10.90	  f2(I30, I31, I32, I33, I34) -> f5(I30, I31, I32, I35, I34) [I35 = I35]
11.02/10.90	  f4(I36, I37, I38, I39, I40) -> f2(I36, I37, I38, I39, I40) 
11.02/10.90	  f2(I41, I42, I43, I44, I45) -> f4(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	  f2(I47, I48, I49, I50, I51) -> f3(I52, I48, I49, I53, I51) [I52 = I52 /\ I51 <= 0 /\ 0 <= I53 /\ I53 <= 0 /\ I53 = I53]
11.02/10.90	  f1(I54, I55, I56, I57, I58) -> f2(I54, I55, I56, I57, I58) 
11.02/10.90	
11.02/10.90	The dependency graph for this problem is:
11.02/10.90	  0 -> 9
11.02/10.90	  1 -> 4, 6, 8
11.02/10.90	  2 -> 1
11.02/10.90	  3 -> 2
11.02/10.90	  4 -> 3
11.02/10.90	  5 -> 
11.02/10.90	  6 -> 5
11.02/10.90	  7 -> 4, 6, 8
11.02/10.90	  8 -> 7
11.02/10.90	  9 -> 4, 6, 8
11.02/10.90	Where:
11.02/10.90	  0) f10#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
11.02/10.90	  1) f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
11.02/10.90	  2) f8#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  3) f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  4) f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  5) f5#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) [I26 = I26]
11.02/10.90	  6) f2#(I30, I31, I32, I33, I34) -> f5#(I30, I31, I32, I35, I34) [I35 = I35]
11.02/10.90	  7) f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 
11.02/10.90	  8) f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	  9) f1#(I54, I55, I56, I57, I58) -> f2#(I54, I55, I56, I57, I58) 
11.02/10.90	
11.02/10.90	We have the following SCCs.
11.02/10.90	  { 1, 2, 3, 4, 7, 8 }
11.02/10.90	
11.02/10.90	DP problem for innermost termination.
11.02/10.90	P =
11.02/10.90	  f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
11.02/10.90	  f8#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 
11.02/10.90	  f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	R = 
11.02/10.90	  f10(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
11.02/10.90	  f9(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
11.02/10.90	  f8(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  f7(I10, I11, I12, I13, I14) -> f8(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f6(I20, I21, I22, I23, I24) -> f3(rnd1, I21, I22, I23, -1 + I24) [rnd1 = rnd1 /\ -1 + I24 <= 0]
11.02/10.90	  f5(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) [I26 = I26]
11.02/10.90	  f2(I30, I31, I32, I33, I34) -> f5(I30, I31, I32, I35, I34) [I35 = I35]
11.02/10.90	  f4(I36, I37, I38, I39, I40) -> f2(I36, I37, I38, I39, I40) 
11.02/10.90	  f2(I41, I42, I43, I44, I45) -> f4(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	  f2(I47, I48, I49, I50, I51) -> f3(I52, I48, I49, I53, I51) [I52 = I52 /\ I51 <= 0 /\ 0 <= I53 /\ I53 <= 0 /\ I53 = I53]
11.02/10.90	  f1(I54, I55, I56, I57, I58) -> f2(I54, I55, I56, I57, I58) 
11.02/10.90	
11.02/10.90	We use the basic value criterion with the projection function NU:
11.02/10.90	  NU[f4#(z1,z2,z3,z4,z5)] = z5
11.02/10.90	  NU[f7#(z1,z2,z3,z4,z5)] = z5
11.02/10.90	  NU[f8#(z1,z2,z3,z4,z5)] = z5
11.02/10.90	  NU[f2#(z1,z2,z3,z4,z5)] = z5
11.02/10.90	  NU[f9#(z1,z2,z3,z4,z5)] = z5
11.02/10.90	
11.02/10.90	This gives the following inequalities:
11.02/10.90	    ==>  I4 (>! \union =) I4
11.02/10.90	  0 <= -1 + -1 + I9  ==>  I9 >! -1 + I9
11.02/10.90	  I12 = I12  ==>  I14 (>! \union =) I14
11.02/10.90	  rnd4 = rnd4  ==>  I19 (>! \union =) I19
11.02/10.90	    ==>  I40 (>! \union =) I40
11.02/10.90	  0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46  ==>  I45 (>! \union =) I45
11.02/10.90	
11.02/10.90	We remove all the strictly oriented dependency pairs.
11.02/10.90	
11.02/10.90	DP problem for innermost termination.
11.02/10.90	P =
11.02/10.90	  f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
11.02/10.90	  f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 
11.02/10.90	  f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	R = 
11.02/10.90	  f10(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
11.02/10.90	  f9(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
11.02/10.90	  f8(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  f7(I10, I11, I12, I13, I14) -> f8(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f6(I20, I21, I22, I23, I24) -> f3(rnd1, I21, I22, I23, -1 + I24) [rnd1 = rnd1 /\ -1 + I24 <= 0]
11.02/10.90	  f5(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) [I26 = I26]
11.02/10.90	  f2(I30, I31, I32, I33, I34) -> f5(I30, I31, I32, I35, I34) [I35 = I35]
11.02/10.90	  f4(I36, I37, I38, I39, I40) -> f2(I36, I37, I38, I39, I40) 
11.02/10.90	  f2(I41, I42, I43, I44, I45) -> f4(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	  f2(I47, I48, I49, I50, I51) -> f3(I52, I48, I49, I53, I51) [I52 = I52 /\ I51 <= 0 /\ 0 <= I53 /\ I53 <= 0 /\ I53 = I53]
11.02/10.90	  f1(I54, I55, I56, I57, I58) -> f2(I54, I55, I56, I57, I58) 
11.02/10.90	
11.02/10.90	The dependency graph for this problem is:
11.02/10.90	  1 -> 4, 8
11.02/10.90	  3 -> 
11.02/10.90	  4 -> 3
11.02/10.90	  7 -> 4, 8
11.02/10.90	  8 -> 7
11.02/10.90	Where:
11.02/10.90	  1) f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
11.02/10.90	  3) f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  4) f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  7) f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 
11.02/10.90	  8) f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	
11.02/10.90	We have the following SCCs.
11.02/10.90	  { 7, 8 }
11.02/10.90	
11.02/10.90	DP problem for innermost termination.
11.02/10.90	P =
11.02/10.90	  f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 
11.02/10.90	  f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	R = 
11.02/10.90	  f10(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
11.02/10.90	  f9(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
11.02/10.90	  f8(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9]
11.02/10.90	  f7(I10, I11, I12, I13, I14) -> f8(I10, I11, I12, I13, I14) [I12 = I12]
11.02/10.90	  f2(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, rnd4, I19) [rnd4 = rnd4]
11.02/10.90	  f6(I20, I21, I22, I23, I24) -> f3(rnd1, I21, I22, I23, -1 + I24) [rnd1 = rnd1 /\ -1 + I24 <= 0]
11.02/10.90	  f5(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) [I26 = I26]
11.02/10.90	  f2(I30, I31, I32, I33, I34) -> f5(I30, I31, I32, I35, I34) [I35 = I35]
11.02/10.90	  f4(I36, I37, I38, I39, I40) -> f2(I36, I37, I38, I39, I40) 
11.02/10.90	  f2(I41, I42, I43, I44, I45) -> f4(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46]
11.02/10.90	  f2(I47, I48, I49, I50, I51) -> f3(I52, I48, I49, I53, I51) [I52 = I52 /\ I51 <= 0 /\ 0 <= I53 /\ I53 <= 0 /\ I53 = I53]
11.02/10.90	  f1(I54, I55, I56, I57, I58) -> f2(I54, I55, I56, I57, I58) 
11.02/10.90	
11.02/13.87	EOF