{1} 1. ~A ∨ (A → B) Prem. {2} 2. A Assum. {3} 3. ~A Assum. (1st D) {4} 4. ~B Assum. {3,4} 5. ~A & ~B 3,4 &I {3,4} 6. ~A 5 &E {2,3,4} 7. A & ~A 2,6 &I {2,3} 8. B 4,7 RAA (1st C) {9} 9. A → B Assum. (2nd D) {2,9} 10. B 2,9 MP (2nd C) {1,2} 11. B 1,3,8,9,10 ∨E {1} 12. A → B 2,311 CP
The best way to prove an implication is usually to start by assuming the antecedent of the implication to be concluded (line 2). To eliminate the disjunction, I also have to assume each of the disjuncts (lines 3 and 9). The first disjunct contradicts the assumption in line 2, and given a contradiction, you can essentially conclude whatever you want. I needed to conclude B, but it should be noted that I didn't get that conclusion for free; it depends on the assumptions of lines 2 and 3 which will have to be eliminated (line 8). The second disjunct also lets me conclude B using modus ponens. Since both disjuncts imply the same conclusion of B (line 11), I can eliminate the assumptions of lines 3 and 9. The assumption on line 2 led to the conclusion on line 11, so I can eliminate the initial assumptionit and conclude that A → B.