That's the law of composition:
That's a basic law, so we prove it with a truth table:
(A → B) & (A → C) ├ A→ (B & C)
Place TTTTFFFF under the A's, TTFFTTFF under the B's and TFTFTFTF under
the C's:
(A → B) & (A → C) ├ A→ (B & C)
T T T T T T T
T T T F T T F
T F T T T F T
T F T F T F F
F T F T F T T
F T F F F T F
F F F T F F T
F F F F F F F
Under the first → put F only when it has T on the left and F on the
right. Otherwise put T. Then you erase what's under the A and the B.
(A → B) & (A → C) ├ A→ (B & C)
T T T T T T
T T F T T F
F T T T F T
F T F T F F
T F T F T T
T F F F T F
T F T F F T
T F F F F F
Under the second →, do the same. Put F only when it has T on the left
and F on the right. Otherwise put T. Then you erase what's under the
A and the C.
(A → B) & (A → C) ├ A→ (B & C)
T T T T T
T F T T F
F T T F T
F F T F F
T T F T T
T T F T F
T T F F T
T T F F F
Under the first & put T only when it has T's on both sides of the &, otherwise
put F, then erase what's under the first two →'s
(A → B) & (A → C) ├ A→ (B & C)
T T T T
F T T F
F T F T
F T F F
T F T T
T F T F
T F F T
T F F F
Under the second & put T only when it has T's on both sides of the &, otherwise
put F, then erase what's under the B and C:
(A → B) & (A → C) ├ A → (B & C)
T T T
F T F
F T F
F T F
T F T
T F F
T F F
T F F
Under the third →, do the same as before. Put F only when it has T on
the left and F on the right. Otherwise put T. Then you erase what's
under the A and the (B & C).
(A → B) & (A → C) ├ A → (B & C)
T T
F F
F F
F F
T T
T T
T T
T T
Finally under the ├, place T is both sides are the same, and put F
otherwise. Then erase the columns on each side of it.
(A → B) & (A → C) ├ A → (B & C)
T
T
T
T
T
T
T
T
Since they are all T, the law of composition is proved.
Edwin
