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[PDF File]Dave’s CPSC 121 Tutorial Notes – Week Two Cheat …
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Transitivity: p → q Proof by cases: p → r (Hypothetical syllogism): q → r q → r ∴ p → r ∴ (p∨q) → r Resolution: p∨q ∼p∨r ∴ (q ∨r) • Multiplexer (Selector) Logic: Variable s is to select between variables p and q: If s is true then be equal to p, otherwise (s is …
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[PDF File]p q l p q p q p r q p r q l q l p - CS Department
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[( p q) [ p o (q o r] o r Rule of Conditional Proof 10. [( p o r) (q o r)]o [( p q) o r] Rule for Proof by Cases 11. [( p o q) (r o s) ( p r)] o (q s) Rule of the Constructive Dilemma 12. [( p o q) (r o s) ( q s)] o ( p r ) Rule of the Destructive Dilemma 13.
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[PDF File]Propositional Logic
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p ^ q ! r means (p ^ q ) ! r When in doubt, use parenthesis. c Xin He (University at Buffalo) CSE 191 Discrete Structures 19 / 37 Translating logical formulas to English sentences Using the above logic operators, we can construct more complicated logical formulas. (They are calledcompound propositions.) Example Proposition p : Alice is smart.
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[PDF File]TruthTables,Tautologies,andLogicalEquivalences
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For example, the compound statement P → (Q∨ ¬R) is built using the logical connectives →, ∨, and ¬. The truth or falsity of P → (Q∨ ¬R) depends on the truth or falsity of P, Q, and R. A truthtableshows how the truth or falsity of a compound statement depends on the truth or falsity of the simple statements from which it’s ...
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[PDF File]CMSC 203 Section 0201: Homework1 Solution
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p q ~p p V q ~p ^ q (p V q) V (~p ^ q) (p V q) V (~p ^ q) → q T T F T F T T T F F T F T F F T T T T T T F F T F F F T Problem 18: (15 points) Write each of the following three statements in the symbolic form and determine which pairs are logically equivalent a.
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[PDF File]BasicArgumentForms - Colorado State University
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if p then q; and if r then s; but either not q or not s; therefore either not p or not r Simplišcation (p∧q) ∴ p p and q are true; therefore p is true Conjunction p,q ∴ (p∧q) p and q are true separately; therefore they are true conjointly Addition p ∴ (p∨q) p is true; therefore the disjunction (p or q) is …
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[PDF File]EJERCICIOS RESUELTOS – 6
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p ∨ (p → q ∧ r) p ∨ (p → q ∧ r) p=V p=F V ∨ (V → q ∧ r) F ∨ (F → q ∧ r) V F ∨ V V La fbf no es contingente, ya que resulta V en todas las interpretaciones (y F en ningu-na). EJERCICIO 6.19 Usando el método de resolución veritativo-funcional, comprobar si las siguientes fbfs
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[PDF File]Truth Tables - Weatherford College
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7. Determine the truth value for ~p Ʌ (~q V r) when p is false, q is true, and r is false. ~p Ʌ (~q V r) Original statement ~F Ʌ (~T V F) Original statement with truth values ~F Ʌ (F V F) Perform the negation in the parenthesis ~F Ʌ (F) Finish the parenthesis T Ʌ (F) Perform the negation False Perform the conjunction 8.
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[PDF File]2. Propositional Equivalences 2.1. Tautology/Contradiction ...
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Example 2.1.3. p_q!:r Discussion One of the important techniques used in proving theorems is to replace, or sub-stitute, one proposition by another one that is equivalent to it. In this section we will list some of the basic propositional equivalences and show how they can be used to prove other equivalences.
negation of p implies q

[PDF File]1.3 Propositional Equivalences
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ICS 141: Discrete Mathematics I (Fall 2014) 1.3 Propositional Equivalences Tautologies, Contradictions, and Contingencies A tautology is a compound proposition which is always true.
p or q implies r

[PDF File]1.3 Propositional Equivalences
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ICS 141: Discrete Mathematics I (Fall 2014) 1.3 Propositional Equivalences Tautologies, Contradictions, and Contingencies A tautology is a compound proposition which is always true.
negation of p q

[PDF File]Dave’s CPSC 121 Tutorial Notes – Week Two Cheat Sheet Tips
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Transitivity: p → q Proof by cases: p → r (Hypothetical syllogism): q → r q → r ∴ p → r ∴ (p∨q) → r Resolution: p∨q ∼p∨r ∴ (q ∨r) • Multiplexer (Selector) Logic: Variable s is to select between variables p and q: If s is true then be equal to p, otherwise (s is false) then be equal to q: (s∧p)∨(∼s∧q) 1
p q r table

[PDF File]BasicArgumentForms - Colorado State University
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if p then q; and if r then s; but either not q or not s; therefore either not p or not r Simplišcation (p∧q) ∴ p p and q are true; therefore p is true ...
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P q p r p q r