11.03/10.90	YES
11.03/10.90	
11.03/10.90	DP problem for innermost termination.
11.03/10.90	P =
11.03/10.90	  f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4#(I37, I38, I39, I40, I41, I42, I43) -> f5#(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4#(I44, I45, I46, I47, I48, I49, I50) -> f5#(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3#(I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	R = 
11.03/10.90	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0]
11.03/10.90	  f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34]
11.03/10.90	  f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	The dependency graph for this problem is:
11.03/10.90	  0 -> 1
11.03/10.90	  1 -> 3
11.03/10.90	  2 -> 3, 4, 5
11.03/10.90	  3 -> 2
11.03/10.90	  4 -> 
11.03/10.90	  5 -> 
11.03/10.90	  6 -> 3, 4, 5
11.03/10.90	  7 -> 2
11.03/10.90	Where:
11.03/10.90	  0) f9#(x1, x2, x3, x4, x5, x6, x7) -> f8#(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  1) f8#(I0, I1, I2, I3, I4, I5, I6) -> f4#(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  3) f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  4) f4#(I37, I38, I39, I40, I41, I42, I43) -> f5#(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  5) f4#(I44, I45, I46, I47, I48, I49, I50) -> f5#(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  6) f3#(I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  7) f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	We have the following SCCs.
11.03/10.90	  { 2, 3 }
11.03/10.90	
11.03/10.90	DP problem for innermost termination.
11.03/10.90	P =
11.03/10.90	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4#(I14, I15, I16, I17, I18, I19, I20) -> f2#(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	R = 
11.03/10.90	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0]
11.03/10.90	  f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34]
11.03/10.90	  f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	We use the reverse value criterion with the projection function NU:
11.03/10.90	  NU[f4#(z1,z2,z3,z4,z5,z6,z7)] = -1 + z4 - z5 + -1 * 0
11.03/10.90	  NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = -1 + z4 - (1 + z5) + -1 * 0
11.03/10.90	
11.03/10.90	This gives the following inequalities:
11.03/10.90	    ==>  -1 + I10 - (1 + I11) + -1 * 0 >= -1 + I10 - (1 + I11) + -1 * 0
11.03/10.90	  I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1  ==>  -1 + I17 - I18 + -1 * 0 > -1 + I17 - (1 + I18) + -1 * 0  with  -1 + I17 - I18 + -1 * 0 >= 0
11.03/10.90	
11.03/10.90	We remove all the strictly oriented dependency pairs.
11.03/10.90	
11.03/10.90	DP problem for innermost termination.
11.03/10.90	P =
11.03/10.90	  f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	R = 
11.03/10.90	  f9(x1, x2, x3, x4, x5, x6, x7) -> f8(x1, x2, x3, x4, x5, x6, x7) 
11.03/10.90	  f8(I0, I1, I2, I3, I4, I5, I6) -> f4(I0, I1, I2, 400, 0, 0, I6) 
11.03/10.90	  f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	  f4(I14, I15, I16, I17, I18, I19, I20) -> f2(rnd1, I20, I16, I17, I18, rnd6, I20) [I19 <= 0 /\ 0 <= I19 /\ 0 <= -1 + I17 - I18 /\ y1 = I20 /\ rnd6 = y1 /\ rnd1 = rnd1]
11.03/10.90	  f4(I21, I22, I23, I24, I25, I26, I27) -> f7(I28, I22, I26, I24, I25, I26, I27) [I28 = I26 /\ I24 - I25 <= 0 /\ 0 <= I26 /\ I26 <= 0]
11.03/10.90	  f5(I29, I30, I31, I32, I33, I34, I35) -> f6(I36, I30, I34, I32, I33, I34, I35) [I36 = I34]
11.03/10.90	  f4(I37, I38, I39, I40, I41, I42, I43) -> f5(I37, I38, I39, I40, I41, I42, I43) [1 <= I42]
11.03/10.90	  f4(I44, I45, I46, I47, I48, I49, I50) -> f5(I44, I45, I46, I47, I48, I49, I50) [1 + I49 <= 0]
11.03/10.90	  f3(I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
11.03/10.90	  f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I68, I67, I63, I64, I65, I69, I67) [I70 = I67 /\ I69 = I70 /\ I68 = I68]
11.03/10.90	
11.03/10.90	The dependency graph for this problem is:
11.03/10.90	  2 -> 
11.03/10.90	Where:
11.03/10.90	  2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f4#(I7, I8, I9, I10, 1 + I11, I12, I13) 
11.03/10.90	
11.03/10.90	We have the following SCCs.
11.03/10.90	  
11.03/13.87	EOF