Chapter 2 The Logic of Quantified Statements. Section 2.4 Arguments with Quantified Statements.

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Presentation transcript:

Chapter 2 The Logic of Quantified Statements

Section 2.4 Arguments with Quantified Statements

Universal Instantiation Universal instantiation “If some property is true of everything in a domain, then it is true of any particular thing in the domain” All men are mortal. Socrates is man. ∴ Socrates is mortal. – Universal instantiation is the fundamental tool of deductive reasoning.

Example Example: r k+1 * r where r is a particular real number and k is a particular integer. 1.For all real numbers x and all integers m and n, x m * x n = x m+n. 2.For all real numbers x, x 1 = x. – r k+1 * r = r k+1 * r 1 – = r (k+1)+1 – = r k+2

Example r k+1 * r = r k+1 * r 1 – For all real numbers x, x 1 = x. (universal truth) – r is a particular real number. (particular instance) – ∴ r 1 = r. = r (k+1)+1 – For all real numbers x and all integers m and n, x m * x n = x m+n – r is a particular real number and k + 1 and 1 are particular integers. – ∴ r k+1 * r 1 = r (k+1)+1

Universal Modus Ponens Universal modus ponens is a combination of universal instantiation and modus ponens. Universal Modus Ponens – If x makes P(x) true, then x makes Q(x) true. – a makes P(a) true. – ∴ a makes Q(a) true.

Example Rewrite the argument using quantifiers, variables, and predicate symbols. – If a number is even, then its square is even. – k is a particular number that is even. – ∴ k 2 is even – major premise ∀ x, if x is even then x 2 is even. – E(x) be “x is even”, S(x) be “x 2 is even”, and let k stand for a particular number that is even. The argument becomes: ∀ x, if E(x) then S(x). E(k), for a particular k. ∴ S(k) (argument has the form of universal modus ponens)

Example Write the conclusion inferred using universal modus ponens. – If T is any right triangle with hypotenuse c and legs a and b, then c 2 = a 2 + b 2. – The right triangle shown has both legs = 1 and hypotenuse c. – ∴ ____________________

Universal Modus Tollens Universal modus tollens combines universal instantiation with modus tollens. Universal Modus Tollens (formal) – ∀ x, if P(x) then Q(x). – ~Q(a), for a particular a. – ∴ ~P(a)

Example Rewrite the following argument using quantifiers, variables, and predicate symbols. Write the major premise. Is this argument valid. – All human beings are mortal. – Zeus is not mortal. – ∴ Zeus is not human. – major premise ∀ x, if x is human then x is mortal. H(x) be “x is human.”, M(x) be “x is mortal” and Z be Zeus ∀ x, if H(x) then M(x) ~M(Z) ∴ ~H(Z) (argument form of universal modus tollens; valid)

Example Write the conclusion that can be inferred using universal modus tollens. – All professors are absent minded. – Tom Jones is not absent-minded. – ∴ ______________________

Validity of Arguments with Quantified Statements An argument is valid if, and only if, the truth of its conclusion follows necessarily from the truth of its premises.

Diagrams to Test for Validity Example – (informal) All integers are rational numbers. – (formal) ∀ integers n, n is a rational number.

Example Using a diagram show validity – All human beings are mortal. – Zeus is not mortal. – ∴ Zeus is not a human being.

Diagrams for Invalidity Show invalidity of the following argument – All human beings are mortal. – Felix is mortal. – ∴ Felix is a human being.

Diagrams for Invalidity

Converse & Inverse Error Converse Error (Quantified) – ∀ x, if P(x) then Q(x). – Q(a) for a particular a. – ∴ P(a) (invalid conclusion) Inverse Error (Quantified) – If x makes P(x) true, then x makes Q(x) true. – a does not make P(a) true. – ∴ a does not make Q(a) true. (invalid conclusion)

Argument with “No” Example: Use diagrams to test the following argument for validity. – No polynomial functions have horizontal asymptotes. – This function has a horizontal asymptote. – ∴ This function is not a polynomial function.

Argument with “No” Rewrite the previous problem in universal modus tollens (formal form) – P(x) be “x is a polynomial function” – Q(x) be “x does not have a horizontal asymptote” – a be “this function” Formal – ∀ x, if P(x) then Q(x). – ~Q(a), for a particular a. – ∴ ~P(a) (universal modus tollens)

Universal Transitivity Formal Version – ∀ x(P(x) → Q(x)). – ∀ x(Q(x) → R(x)). – ∴∀ x(P(x) → R(x)).

Example Rewrite the informal to formal and show validity through Universal Transitivity. 1.All the triangles are blue. 2.If an object is to the right of all the squares, then it is above all the circles. 3.If an object is not to the right of all the squares, then it is not blue. ∴ All the triangles are above all the circles.

Example 1. ∀ x, if x is a triangle, then x is blue. 2. ∀ x, if x is to the right of all the squares, then x is above all circles. 3. ∀ x, if x is not to the right of all the squares, then x is not blue. ∴ ∀ x, if x is a triangle, then x is above all the circles. Reorder (premises such that the conclusion of each is the hypothesis of the next) 1. ∀ x, if x is a triangle, then x is blue. 3. ∀ x, if x is blue, then x is to the right of all the squares. (contrapositive form, if not q then not p) 2. ∀ x, if x is to the right of all the squares, then x is above all circles. ∴ ∀ x, if x is a triangle, then x is above all the circles.