Thanks, that makes sense. And the added explanation helps a lot, I can see the argument for going from ⊢A→B to A⊢B now:
PA proves A→B;
So PA + A certainly proves A→B, since it can prove everything PA can;
Also, PA + A proves A;
So we have A⊢A∧(¬A∨B);
Which in turn gives us A⊢B.
Thanks, that makes sense. And the added explanation helps a lot, I can see the argument for going from ⊢A→B to A⊢B now:
PA proves A→B;
So PA + A certainly proves A→B, since it can prove everything PA can;
Also, PA + A proves A;
So we have A⊢A∧(¬A∨B);
Which in turn gives us A⊢B.