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