let a, b be non zero Element of NAT ; :: thesis:
reconsider an = a, bn = b as non zero Nat ;
assume A1: a,b are_coprime ; :: thesis:

now :: thesis:

let i be object ; :: thesis:
assume i in SetPrimes ; :: thesis:
then reconsider p = i as prime Element of NAT by NEWTON:def 6;
A2: p |-count (an * bn) = (p |-count a) + (p |-count b) by NAT_3:28;
A3: a gcd b = 1 by A1, INT_2:def 3;
A4: p > 1 by INT_2:def 4;

per cases ;

suppose A5: p |-count (a * b) = 0 ; :: thesis:

then A6: p |-count b = 0 by A2;
A7: p |-count a = 0 by A2, A5;
thus (PFactors (a * b)) . i = 0 by A5, Th48
.= 0 + ((PFactors b) . i) by A6, Th48
.= ((PFactors a) . i) + ((PFactors b) . i) by A7, Th48
.= ((PFactors a) + (PFactors b)) . i by PRE_POLY:def 5 ; :: thesis:

end;

suppose A8: p |-count (a * b) <> 0 ; :: thesis:

thus (PFactors (a * b)) . i = ((PFactors a) + (PFactors b)) . i :: thesis:

proof

per cases by A2, A8;

suppose A9: p |-count a <> 0 ; :: thesis:

A10: now :: thesis:

consider ka being Nat such that
A11: p |-count a = ka + 1 by A9, NAT_1:6;
reconsider ka = ka as Element of NAT by ORDINAL1:def 12;
p |^ (p |-count a) divides a by A4, NAT_3:def 7;
then p * (p |^ ka) divides a by A11, NEWTON:6;
then consider la being Nat such that
A12: a = (p * (p |^ ka)) * la by NAT_D:def 3;
a = p * ((p |^ ka) * la) by A12;
then A13: p divides a by NAT_D:def 3;
assume p |-count b <> 0 ; :: thesis:
then consider kb being Nat such that
A14: p |-count b = kb + 1 by NAT_1:6;
reconsider kb = kb as Element of NAT by ORDINAL1:def 12;
p |^ (p |-count b) divides b by A4, NAT_3:def 7;
then p * (p |^ kb) divides b by A14, NEWTON:6;
then consider lb being Nat such that
A15: b = (p * (p |^ kb)) * lb by NAT_D:def 3;
b = p * ((p |^ kb) * lb) by A15;
then p divides b by NAT_D:def 3;
then p divides 1 by A3, A13, NAT_D:def 5;
hence contradiction by A4, NAT_D:7; :: thesis:

end;

thus (PFactors (a * b)) . i = p by A8, Th49
.= ((PFactors a) . p) + 0 by A9, Th49
.= ((PFactors a) . p) + ((PFactors b) . p) by A10, Th48
.= ((PFactors a) + (PFactors b)) . i by PRE_POLY:def 5 ; :: thesis:

end;

suppose A16: p |-count b <> 0 ; :: thesis:

A17: now :: thesis:

assume p |-count a <> 0 ; :: thesis:
then consider ka being Nat such that
A18: p |-count a = ka + 1 by NAT_1:6;
reconsider ka = ka as Element of NAT by ORDINAL1:def 12;
p |^ (p |-count a) divides a by A4, NAT_3:def 7;
then p * (p |^ ka) divides a by A18, NEWTON:6;
then consider la being Nat such that
A19: a = (p * (p |^ ka)) * la by NAT_D:def 3;
a = p * ((p |^ ka) * la) by A19;
then A20: p divides a by NAT_D:def 3;
consider kb being Nat such that
A21: p |-count b = kb + 1 by A16, NAT_1:6;
reconsider kb = kb as Element of NAT by ORDINAL1:def 12;
p |^ (p |-count b) divides b by A4, NAT_3:def 7;
then p * (p |^ kb) divides b by A21, NEWTON:6;
then consider lb being Nat such that
A22: b = (p * (p |^ kb)) * lb by NAT_D:def 3;
b = p * ((p |^ kb) * lb) by A22;
then p divides b by NAT_D:def 3;
then p divides 1 by A3, A20, NAT_D:def 5;
hence contradiction by A4, NAT_D:7; :: thesis:

end;

thus (PFactors (a * b)) . i = p by A8, Th49
.= 0 + ((PFactors b) . p) by A16, Th49
.= ((PFactors a) . p) + ((PFactors b) . p) by A17, Th48
.= ((PFactors a) + (PFactors b)) . i by PRE_POLY:def 5 ; :: thesis:

end;

end;

end;

end;

end;

end;

hence PFactors (a * b) = (PFactors a) + (PFactors b) by PBOOLE:3; :: thesis: