P q r q p r ∴ q aka Disjunction Elimination Corresponding Tautology ((p q) ∧ (r q) ∧ (p r )) q Example Let p be "I will study discrete math" Let q be "I will study Computer Science" Let r be "I will study databases" "If I will study discrete math, then I will study Computer Science" 2 P → Q A s s u m This second assumption is good too, since you basically want to do a DeMorgan on line 1, and to do those in Natural Deduction, you do a Proof by COntradiction on each of the disjuncts so yes, assume the disjunct, but then proceed with getting the full disjunct 3 ( P → Q) ∨ ( Q → P) ∨ I n t r o ( 2) and now youRule of inference of type PC says that if set B= {p1, p2,, pn} and p is propositional consequence of set B, then p1^p2^^pn→p is tautology Then in this case (B, p) is an order pairs or rule of inference of type PC For your case, let Set B'= {p→q, p} and we know that q is propositional consequence of set B' because p→q^p→q is a tautology
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Divide: p - q p + q p - q-Using the definitions of the connectives in Section 02, we see that for this to be true, either P → Q P → Q must be true or Q→ R Q → R must be true (or both) Those are true if either P P is false or Q Q is true (in the first case) and Q Q is false or R R is true (in the second case) So—yeah, it Given p ⇒ q, use the Fitch System to prove ¬p ∨ q 1 p => q Premise 2 ~(~p q) Assumption 3 ~p Assumption 4
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Two propositions p and q are called logically equivalent if and only if vp = vq holds for all valuations v on Prop In other words, two propositions p and q are logically equivalent if and only if p 㲗 q is a tautology We write p ≡ q if and only if p and q are logically equivalent We have shown that (¬p ⋁q) ≡ (p q) In generalStatement II is true;P = it rains / is raining q = the squirrels hide / are hiding ' 05Œ09, N Van Cleave 1 Rewriting the Premises and Conclusion Premise 1 p → q Premise 2 p Conclusion q Thus, the argument converts to ((p → q) ∧ p) → q With Truth Table p q ((p → q) ∧ p) → q T T T F F T F F
The conditional of q by p is "If p then q " or " p implies q " and is denoted by p q It is false when p is true and q is false;P = "" Q = "" R = "Calvin Butterball has purple socks" I want to determine the truth value of Since I was given specific truth values for P, Q, and R, I set up a truth table with a single row using the given values for P, Q, and R Therefore, the statement is trueP Q P Q T T T T F F F T F F F T Remember that "P Q" is just shorthand for "(P Q) (Q P)" This is a conjunction (This is because, as we will soon learn the "main operator" in this formula is the " ") But, conjunctions such as "P Q" only comes out true when "P" and "Q" are BOTH true So,
Example 232 Show (p!q) is equivalent to p^q Solution 1 Build a truth table containing each of the statements p q q p!q (p!q) p^q T T F T F F T F T F T T F T F T F F F F T T F F Since the truth values for (p!q) and p^qare exactly the same for all possible combinations of truth values of pand q, the two propositions are equivalentP then q" or "p implies q", represented "p → q" is called a conditional proposition For instance "if John is from Chicago then John is from Illinois" The proposition p is called hypothesis or antecedent, and the proposition q is the conclusion or consequent Note that p → q is true always except when p is true and q is falseNow, our final goal is to be able to fill in truth tables with more compound statements which have more than just one logical connective in them Statements like q→~s or (r∧~p)→r or (q&rarr~p)∧(p↔r) have multiple logical connectives, so we will need to do them one step at a time using the order of operations we defined at the beginning of this lecture
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A biconditional statement is really a combination of a conditional statement and its converse The biconditional operator is denoted by a doubleheaded arrow P ↔ Q {P \leftrightarrow Q} P↔Q is read as " P if and(p^q) p_q p q (p^q) !9 rows (pVq) V (~p^q) → q p q ~p p V q ~p ^ q (p V q) V (~p ^ q) (p V q) V (~p ^ q) → q T T
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∴ ((p∨q)∧(p∨r)) p or (q and r) is equiv to (p or q) and (p or r) Double Negation p ∴ ¬¬p p is equivalent to the negation of not p Transposition (p → q) ∴ (¬q → ¬p) if p then q is equiv to if not q then not p Material Implication (p → q) ∴ (¬p∨q) if p then q is equiv to not p or q Exportation ((p∧q) → r) ∴ (p → (q → r))P → q = (~p ∨ q) In the Principia Mathematica, the "=" denotes "is defined to mean" Using this denotation, the above expression can be read "p implies q is defined to mean that either p is false or q is true" The following truth table shows the logical equivalence of "If pOtherwise it is true Contrapositive The contrapositive of a conditional statement of the form "If p then q " is "If ~ q then ~ p " Symbolically, the contrapositive of
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Statement − I (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy Statement − II (p → q) ↔ (~ q → ~ p) is a tautology Statement I is True;P q ~q (p ↔ ~q) ~ (p ↔ ~q) p ↔ q T T F F T T T F T T F F F T F T F F F F T F T T Clearly, ~ (p ↔ ~q) is not a tautology because it does not contain T in the column of its truth table Also, ~(p ↔ ~ q) & p ↔ q have the same truth valueP → q (p implies q) (if p then q) is the proposition that is false when p is true and q is false and true otherwise Equivalent to finot p or qfl Ex If I am elected then I will lower the taxes If you get 100% on the final then you will get an A p I am elected q I will lower the taxes Think of it as a contract, obligation or pledge
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P q p Ʌ T T T T F F F T F F F F →IV Conditional ( ifthen)is false only V Biconditional (↔ if and only if)is st when the antecedent (1 ) is true and true only when the component ndthe component (2 ) is false statements have the same truth value p q p → q👉 Learn how to use the Rational Zero Test on Polynomial expression Rational Zero Test or Rational Root test provide us with a list of all possible real ZerStatementII is a correct explanation for StatementI Statement I is True;
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The basic method I would use is to use P>Q ~P V Q, or prove it using truth tables Then use boolean algebra with DeMorgan's law to make the right side of your equation equivalent to the left sideP _ q T T F T T F F F T F T F T F T T F T T F T T T F F F T F T T T T Since (p ^q) !p _q is T in all cases, therefore (p^q) p_q You could stop one step earlier by noticing that since the columns for (p ^q) and p _q are identical, therefore they're logically equivalent 12Solution The compound statement (p q) p consists of the individual statements p, q, and p q The truth table above shows that (p q) p is true regardless of the truth value of the individual statements Therefore, (p q) p is a tautology In the examples below, we will determine whether the given statement is a tautology by creating a truth table
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Two propositions p and q arelogically equivalentif their truth tables are the same Namely, p and q arelogically equivalentif p $ q is a tautology If p and q are logically equivalent, we write p q c Xin He (University at Buffalo) CSE 191 Discrete Structures 22 / 37PJohn is a student qUKisauniversity Compound statement astatement that is formed of primitive statements with logical connectives such as 1 Negation p (or,¬p) 2 Conjunction p Λq (p and q) 3 Disjunction p V q (p or q) 4 Implication p →q (p implies q)To check if $\neg (p \vee q)$ and $\neg p \wedge \neg q$ are logically equivalent \begin{array}{ccccccc} p & q & p \vee q & \neg (p \vee q) & \neg p & \neg q & \neg p \wedge \neg q \\\hline T & T & T & F & F & F & F \\ T & F & T & F & F & T & F \\ F & T & T & F & T & F & F \\ F & F & F & T & T & T & T \\ \end{array} Since columns
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Conditional (p =)q) ()(˘p_q) ˘(p =)q) ()(p^˘q) Rules of Inference Modus Ponens p =)q Modus Tollens p =)q p ˘q) q )˘p Elimination p_q Transitivity p =)q ˘q q =)r) p ) p =)r Generalization p =)p_q Specialization p^q =)p q =)p_q p^q =)q Conjunction p Contradiction Rule ˘p =)F q ) p) p^q « 11 BEShapiro forintegraltablecomP → q ≡ ¬(p ∧ ¬q) By De Morgan's laws p → q ≡ ¬(p ∧ ¬q) p → q ≡ ¬p ∨ ¬¬q p → q ≡ ¬p ∨ q Thus p → q ≡ ¬p ∨ q If p is false, then ¬p ∨ q is true If p is true, then q has to be true for the whole expression to be true If p is false, then ¬p ∨ q is true If p is true, then q has to be true for the wholeSteps for Solving Linear Equation P Q = P Q P Q = P − Q Subtract P from both sides Subtract P from both sides PQP=Q P Q − P = − Q Combine all
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Tabel kebenaran untuk ekspresi p → q dengan ~p v q Karena untuk tiaptiap baris, nilai kebenaran pada kolom p → q dan ~p v q sama, maka disimpulkan bahwa p → q ≡ ~p v q Contoh buktikan ekuivalensi tanpa menggunakan tabel kebenaran dari (q → p) ≡ (~p → ~q), karena persamaan kanan lebih komplek maka kita turunkanAnswer (1 of 2) Question originally answered What is the truth table for (p>q) ^ (q>r)> (p>r)?P^q(p^q) p_q "Jan is not rich, or not happy" b)Mei walks or takes the bus to class p = "Mei walks to class" q = Mei takes the bus to class" p_q(p_q) p^q "Mei does not walk to class, and Mei does not take the bus to class" 13 pg 35 # 11 Show that each conditional statement is a tautology without using truth tables b p !(p_q)
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This sentence is of the form "If p then q" So, the symbolic form is p → q wherep You are intelligent q You will pass the exam Converse StatementIf you will pass the exam, then you are intelligent Inverse StatementIf you are not intelligent, then you will not pass the examP(x) =)Q(x) For a particular s 2S, Q(s)) P(s) • Denying the Hypothesis(The conjunction p q r is then interpreted to mean that p, q, and r are all true, p ∨ q ∨ r to mean that at least one of p, q, and r is true, and so forth) Validity in PC Given the standard interpretation, a wff of PC becomes a sentence, true or false, when all its variables are replaced by actual sentences Such a wff is therefore
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Statement II is True;I construct the truth table for (P → Q)∨ (Q→ P) and show that the formula is always true P Q P → Q Q→ P (P → Q)∨ (Q→ P) T T T T T T F F T T F T T F T F F T T T The last column contains only T's Therefore, the formula is a tautology Example Construct a truth table for (P → Q)∧ (Q→ R) P Q R P → Q Q→ R (P → Q)∧ (Q→ R)StatementII is not a correct explanation for StatementI
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Fallacies • A rming the Conclusion p =)q q) p • Universal AC 8x 2S;\begin{array}{cccccccccccccccccc}p&q&r&p \supset q&q\supset r144 Hardegree, Symbolic Logic (MT) P → Q ~Q –––––– ~P This argument form is traditionally called modus tollens, which is short for modus tollendo tollens, which is a Latin expression meaning the mode of denying by denying It is so called because, in this mode of reasoning, one goes from a
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Let's construct a truth table for p v ~q This is read as "p or not q" Step 1 Make a table with different possibilities for p and q There are 4 different possibilities Case 4An expression involving logical variables that is false for all values is called a contradiction Statements that are not tautologies or contradictions are called contingencies Definition 213 We say two propositions p p and q q are logically equivalent if p ↔ q p ↔ q is a tautology We denote this by p ≡ q p ≡ qSince (p→ q)∧(q→ r) → (p→ r) is always T, it is a tautology (0 points) (c) sol p q p→ q p∧(p→ q) p∧(p→ q) → q T T T T T T F F F T F T T F T F F T F T Since p∧(p→ q) → qis always T, it is a tautology (0 points) (d) p q r p∨q p→ r q→ r (p∨q)∧(p→ r)∧(q→ r) (p∨q)∧(p→ r)∧(q→ r) → r
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The p or q proposition is only false if both component propositions p and q are false What is the symbol of Biconditional?Composition of Functions In addition to adding, subtracting, multplying and dividing, two functions can be composed The composition of a function is when the xvalue is replaced by a function For example if p (x) = x 3 and q (x) = x 1, the compostition of p with q is The notation p ∘ q, reads "p composed with q"P ∧ q __ a It is after 5 PM and __ p ∧ ~ q _ __ b It is after 5 PM and they are working they are not working Common English Expressions for p^q Or Statements __Disjunction__ is a compound statement formed using the inclusive orrepresented by the symbol ∨ Thus, "p or q or both" is symbolized by p ∨ q
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Apply the distributive property by multiplying each term of pq by each term of 2p5 2p^{2}5p2qp5q\left(p^{2}3pqp3q\right) Apply the distributive property by multiplying each term of pq by each term of p3
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