8. Propositional Logic Natural deduction - negation. Solved problems

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1 8. Propositional Logic Natural deduction - negation Solved problems

2 Problem: A B can be derived from (A B). (De Morgan law)

3 Problem: A B can be derived from (A B). (De Morgan law) Let us first think intuitively why A B should follow from (A B).

4 Problem: A B can be derived from (A B). (De Morgan law) Let us first think intuitively why A B should follow from (A B). Say, it is not true that it rains or snows. Why can we conclude that it neither rains nor snows? Well, because if it for example rained, then it would a fortiori rain or snow, so we would contradict the underlined assumption.

5 Problem: A B can be derived from (A B). (De Morgan law) Let us first think intuitively why A B should follow from (A B). Say, it is not true that it rains or snows. Why can we conclude that it neither rains nor snows? Well, because if it for example rained, then it would a fortiori rain or snow, so we would contradict the underlined assumption. We try to make this formal.

6 Problem: A B can be derived from (A B). (De Morgan law)

7 Problem: A B can be derived from (A B). (De Morgan law) Let us continue thinking intuitively why A B should follow from (A B).

8 Problem: A B can be derived from (A B). (De Morgan law) Let us continue thinking intuitively why A B should follow from (A B). Since we are proving a conjunction we can take each conjunct separately. Let us look at A. Assuming A gives A B, contradicting immediately the assumption (A B). So we must conclude A. Similarly we get B.

9 Problem: A B can be derived from (A B). (De Morgan law) [A] (A B) A B (A B) (A B) I [B] A B A B

10 Problem: A B can be derived from (A B). (Another de Morgan law) This is more difficult! We want to conclude A B, so the temptation is to try to derive one of A and B. But which one?? This is related to the difference between so called constructive logic and classical logic. Our logic is classical.

11 Problem: A B can be derived from (A B). (Another de Morgan law) Let us first think intuitively why A B should follow from (A B). Say, a dish does not contain both cream and meat. Why can we conclude that either cream is missing or meat is missing? Well, because if both cream and meat were there, we would contradict the underlined assumption, so one of them must be missing. We try to make this formal.

12 Problem: A B can be deri ed from (A B). (Another de Morgan law) Let us again first think intuitively why A B should follow from (A B). Let us assume A B is false i.e. ( A B) and work towards a contradiction. Now clearly A leads to a contradiction, so A i.e. A. Respecti ely B. So A B. This contradicts the assumption (A B). So we get ( A B) i.e. A B.

13 Problem: A B can be derived from (A B). (Another de Morgan law) [ A] [ ( A B)] A B ( A B) ( A B) A A E A B I [ B] B (A B) (A B) ( A B) A B (A B) I E

14 Example: A A is derivable Intuition: We use indirect proof. So we assume (A A) and derive a contradiction. Now A leads to A A and hence to a contradiction. Thus we may conclude A. But this leads to A A and hence to a contradiction, and we are done.

15 Example: A A is derivable We make an assumption A A A

16 Example: A A is derivable A (A A) A A (A A) (A A) Temporary assumption

17 Example: A A is derivable (A A) This assumption is now eliminated [A] A A A (A A) (A A) I

18 Example: A A is derivable (A A) [A] A A A A A (A A) (A A) E

19 Example: A A is derivable [A] (A A) A A (A A) (A A) A (A A) A A (A A) (A A) E

20 Example: A A is derivable This assumption is now eliminated [ (A A)] (A A) [A] A A (A A) (A A) A [ (A A)] A A (A A) (A A) I E

21 Example: A A is derivable [A] [ (A A)] A A (A A) (A A) A [ (A A)] A A (A A) (A A) (A A) A A E I E

MAT25 LECTURE 10 NOTES. = a b. > 0, there exists N N such that if n N, then a n a < ɛ

MAT25 LECTURE 10 NOTES. = a b. > 0, there exists N N such that if n N, then a n a < ɛ MAT5 LECTURE 0 NOTES NATHANIEL GALLUP. Algebraic Limit Theorem Theorem : Algebraic Limit Theorem (Abbott Theorem.3.3) Let (a n ) and ( ) be sequences of real numbers such that lim n a n = a and lim n =