1) Give a general proof to show that for all sentences X and Y and for every consistent set Σ of sentences, if X and Y are logical consequences of Σ, then the set {X, Y} is consistent.
2) Let C be a logical consequence of the set {P1, P2, P3} and of the set {P1, P2, not-P3}. Prove that C is a logical consequence of {P1, P2}.
To prove both statements, we can use the method of proof by contradiction. Here is the general proof for the first statement:
1) Assume {X, Y} is inconsistent, i.e., it leads to a contradiction.
2) Since it leads to a contradiction, any sentence Z can be derived from the inconsistent set {X, Y}.
3) Now, assume that X and Y are logical consequences of Σ. This means that for any model M that satisfies Σ, M also satisfies X and Y.
4) Since X is a logical consequence of Σ, X must be true in any model that satisfies Σ, including the model that led to the contradiction in step 2.
5) Similarly, Y is also true in the same model.
6) But since X and Y are both true in the model that led to the contradiction, {X, Y} cannot be inconsistent. This contradicts our assumption in step 1.
7) Therefore, if X and Y are logical consequences of Σ, then the set {X, Y} must be consistent.
Now let's move on to the second statement:
1) Assume C is not a logical consequence of {P1, P2}.
2) If C is not a logical consequence, then there exists a model M that satisfies {P1, P2} but makes C false.
3) Now, since C is a logical consequence of {P1, P2, P3}, we know that in any model that satisfies {P1, P2, P3}, C must be true.
4) Similarly, since C is a logical consequence of {P1, P2, not-P3}, in any model that satisfies {P1, P2, not-P3}, C must also be true.
5) Now, consider the model M that satisfies {P1, P2} as assumed in step 2. Since M satisfies {P1, P2}, M must also satisfy {P1, P2, not-P3}.
6) But if M satisfies {P1, P2, not-P3}, then C must be true in M, contradicting our assumption in step 2.
7) Therefore, our assumption in step 1 must be false. Hence, C is indeed a logical consequence of {P1, P2}.