**Date:** April 12 2023

**Summary:** TODO

**Keywords:** #archive #proofs #logic #discrete #mathematics #deductive #reasoning

TODO

- Deductive Reasoning
- Premises
- Connective Symbols
- Conjunction, Disjunction, and Negation
- Well-Formed Formulas
- Truth Tables
- Equivalent Formulas
- Tautologies
- Contradictions
- Variables
- Free Variables
- Bound Variables
- Sets
- Truth Set
- Elementhood Test
- Set Operations
- Not Sure Section
- Conditional Statements
- Converses and Contrapositives
- Biconditional Statements

- How To Cite
- References:
- Discussion:

TODO: Finish creating note for this topic TODO: Create Anki flashcards from note See Note

TODO: Finish creating note for this topic TODO: Create Anki flashcards from note See Note

TODO: Finish creating note for this topic TODO: Create Anki flashcards from note See Note

TODO: Add this theorem somewhere

Theorem 1.4.7. For any sets A and B, (A âˆª B) \ B âŠ† A. Proof. We must show that if something is an element of (A âˆª B) \ B, then it must also be an element of A, so suppose that x âˆˆ (A âˆª B) \ B. This means that x âˆˆ A âˆª B and x âˆˆ B, or in other words x âˆˆ A âˆ¨ x âˆˆ B and x âˆˆ B. But notice that these statements have the logical form P âˆ¨ Q and Â¬Q, and this is precisely the form of the premises of our very first example of a deductive argument in Section 1.1! As we saw in that example, from these premises we can conclude that x âˆˆ A must be true. Thus, anything that is an element of (A âˆª B) \ B must also be an element of A, so (A âˆª B) \ B âŠ† A.

Zelko, Jacob. *Chapter 1: Sentential Logic*. https://jacobzelko.com/04122023013725-sentential-logic. April 12 2023.

CC BY-SA 4.0 Jacob Zelko. Last modified: May 19, 2024.
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