25.80/25.65	YES
25.80/25.65	
25.80/25.65	DP problem for innermost termination.
25.80/25.65	P =
25.80/25.65	  f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 
25.80/25.65	  f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) 
25.80/25.65	  f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	  f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 
25.80/25.65	  f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	R = 
25.80/25.65	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
25.80/25.65	  f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 
25.80/25.65	  f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
25.80/25.65	  f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 
25.80/25.65	  f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37]
25.80/25.65	  f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	The dependency graph for this problem is:
25.80/25.65	  0 -> 1
25.80/25.65	  1 -> 8
25.80/25.65	  2 -> 5
25.80/25.65	  3 -> 5
25.80/25.65	  4 -> 6
25.80/25.65	  5 -> 6
25.80/25.65	  6 -> 7
25.80/25.65	  7 -> 10, 11
25.80/25.65	  8 -> 9
25.80/25.65	  9 -> 7
25.80/25.65	  10 -> 2, 3, 4
25.80/25.65	  11 -> 8
25.80/25.65	Where:
25.80/25.65	  0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 
25.80/25.65	  1) f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) 
25.80/25.65	  2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	  8) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 
25.80/25.65	  9) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  10) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  11) f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	We have the following SCCs.
25.80/25.65	  { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 }
25.80/25.65	
25.80/25.65	DP problem for innermost termination.
25.80/25.65	P =
25.80/25.65	  f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	  f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 
25.80/25.65	  f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	R = 
25.80/25.65	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
25.80/25.65	  f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 
25.80/25.65	  f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
25.80/25.65	  f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 
25.80/25.65	  f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37]
25.80/25.65	  f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	We use the extended value criterion with the projection function NU:
25.80/25.65	  NU[f4#(x0,x1,x2,x3)] = -x1 + x3 + 1
25.80/25.65	  NU[f2#(x0,x1,x2,x3)] = -x1 + x3 + 1
25.80/25.65	  NU[f1#(x0,x1,x2,x3)] = -x1 + x3
25.80/25.65	  NU[f6#(x0,x1,x2,x3)] = -x1 + x3
25.80/25.65	  NU[f7#(x0,x1,x2,x3)] = -x1 + x3
25.80/25.65	  NU[f8#(x0,x1,x2,x3)] = -x1 + x3
25.80/25.65	  NU[f3#(x0,x1,x2,x3)] = -x1 + x3
25.80/25.65	
25.80/25.65	This gives the following inequalities:
25.80/25.65	  1 + I4 <= 0  ==>  -I5 + I7 >= -I5 + I7
25.80/25.65	  1 <= I8  ==>  -I9 + I11 >= -I9 + I11
25.80/25.65	  0 <= I12 /\ I12 <= 0  ==>  -I13 + I15 >= -I13 + I15
25.80/25.65	    ==>  -I17 + I19 >= -I17 + (-1 + I19)
25.80/25.65	    ==>  -I21 + I23 >= -I21 + I23
25.80/25.65	    ==>  -I25 + I27 >= -I25 + I27
25.80/25.65	    ==>  -I29 + I31 + 1 >= -I29 + I31 + 1
25.80/25.65	  1 + I33 <= I35  ==>  -I33 + I35 + 1 > -I33 + I35  with  -I33 + I35 + 1 >= 0
25.80/25.65	  rnd1 = rnd1 /\ 1 + I42 <= I43  ==>  -I41 + I43 >= -I41 + I43
25.80/25.65	  I47 <= I46  ==>  -I45 + I47 >= -(1 + I45) + I47 + 1
25.80/25.65	
25.80/25.65	We remove all the strictly oriented dependency pairs.
25.80/25.65	
25.80/25.65	DP problem for innermost termination.
25.80/25.65	P =
25.80/25.65	  f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	  f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 
25.80/25.65	  f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	R = 
25.80/25.65	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
25.80/25.65	  f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 
25.80/25.65	  f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
25.80/25.65	  f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 
25.80/25.65	  f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37]
25.80/25.65	  f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	The dependency graph for this problem is:
25.80/25.65	  2 -> 5
25.80/25.65	  3 -> 5
25.80/25.65	  4 -> 6
25.80/25.65	  5 -> 6
25.80/25.65	  6 -> 7
25.80/25.65	  7 -> 10, 11
25.80/25.65	  8 -> 
25.80/25.65	  10 -> 2, 3, 4
25.80/25.65	  11 -> 8
25.80/25.65	Where:
25.80/25.65	  2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	  8) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 
25.80/25.65	  10) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  11) f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	We have the following SCCs.
25.80/25.65	  { 2, 3, 4, 5, 6, 7, 10 }
25.80/25.65	
25.80/25.65	DP problem for innermost termination.
25.80/25.65	P =
25.80/25.65	  f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	  f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	R = 
25.80/25.65	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
25.80/25.65	  f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 
25.80/25.65	  f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
25.80/25.65	  f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 
25.80/25.65	  f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37]
25.80/25.65	  f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	We use the extended value criterion with the projection function NU:
25.80/25.65	  NU[f1#(x0,x1,x2,x3)] = -x2 + x3 - 1
25.80/25.65	  NU[f6#(x0,x1,x2,x3)] = -x2 + x3 - 1
25.80/25.65	  NU[f7#(x0,x1,x2,x3)] = -x2 + x3 - 2
25.80/25.65	  NU[f8#(x0,x1,x2,x3)] = -x2 + x3 - 2
25.80/25.65	  NU[f3#(x0,x1,x2,x3)] = -x2 + x3 - 2
25.80/25.65	
25.80/25.65	This gives the following inequalities:
25.80/25.65	  1 + I4 <= 0  ==>  -I6 + I7 - 2 >= -I6 + I7 - 2
25.80/25.65	  1 <= I8  ==>  -I10 + I11 - 2 >= -I10 + I11 - 2
25.80/25.65	  0 <= I12 /\ I12 <= 0  ==>  -I14 + I15 - 2 >= -I14 + I15 - 2
25.80/25.65	    ==>  -I18 + I19 - 2 >= -(-1 + I18) + (-1 + I19) - 2
25.80/25.65	    ==>  -I22 + I23 - 2 >= -(1 + I22) + I23 - 1
25.80/25.65	    ==>  -I26 + I27 - 1 >= -I26 + I27 - 1
25.80/25.65	  rnd1 = rnd1 /\ 1 + I42 <= I43  ==>  -I42 + I43 - 1 > -I42 + I43 - 2  with  -I42 + I43 - 1 >= 0
25.80/25.65	
25.80/25.65	We remove all the strictly oriented dependency pairs.
25.80/25.65	
25.80/25.65	DP problem for innermost termination.
25.80/25.65	P =
25.80/25.65	  f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	R = 
25.80/25.65	  f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 
25.80/25.65	  f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 
25.80/25.65	  f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 
25.80/25.65	  f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 
25.80/25.65	  f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 
25.80/25.65	  f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35]
25.80/25.65	  f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37]
25.80/25.65	  f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43]
25.80/25.65	  f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46]
25.80/25.65	
25.80/25.65	The dependency graph for this problem is:
25.80/25.65	  2 -> 5
25.80/25.65	  3 -> 5
25.80/25.65	  4 -> 6
25.80/25.65	  5 -> 6
25.80/25.65	  6 -> 7
25.80/25.65	  7 -> 
25.80/25.65	Where:
25.80/25.65	  2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0]
25.80/25.65	  3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8]
25.80/25.65	  4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0]
25.80/25.65	  5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 
25.80/25.65	  6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 
25.80/25.65	  7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 
25.80/25.65	
25.80/25.65	We have the following SCCs.
25.80/25.65	  
25.80/28.62	EOF