Discrete Structures

# Lecture Notes, 6 Oct 2010

 Last Update: 6 October 2010 Note: or material is highlighted

### §1.5: Rules of Inference (cont'd)

1. Rules of Inference for Quantifiers:

1. Notation:

1. Let A(t) represent any proposition or propositional function
(even one with a multiple-place predicate)
that contains at least one occurrence of the term t.

2. Let A(t1 := t2) represent any proposition or propositional function in which all free occurrences of term t1
are replaced by (i.e., are assigned the value) term t2.

• E.g., if A(x) is: (P(x) → Q(x,y)),
then A(x := c) is: (P(c) → Q(c,y))

2. The rules:

• Universal Instantiation (UI)

xA(x)
A(x := c)    i.e., A(c)

where c could name any object in the domain

• Universal Generalization (UG)

A(c)
xA(c := x)    i.e., ∀xA(x)

where c must name an arbitrary object in the domain
not any particular or special one
(cf. proofs in geometry)

• Existential Instantiation (EI)

xA(x)
A(x := c)    i.e., A(c)

where c must name an arbitrary object in the domain,
not any particular one

i.e., c must be a brand-new name.

• Existential Generalization (EG)

A(c)
xA(c := x)    i.e., ∃xA(x)

where c names some actual object in the domain that is known to satisfy A.

2. Sample Proof of Validity:

1. Argument:

P1: Every student in 191 is a freshman.
P2: Every freshman will pass the midterm.
P3: Fred is a student in 191.
C : ∴ Some student in 191 will pass the midterm.

2. Syntax & Semantics of Representation:

191(x) = x is a student
Frosh(x) = x is a freshman
Pass(x) = x will pass the midterm
fred = Fred

3. Translation:

P1: ∀x[191(x) → Frosh(x)]
P2: ∀x[Frosh(x) → Pass(x)]
P3: 191(fred)
C : ∴ ∃x[191(x) ∧ Pass(x)]

4. Strategy:

Want to show C (i.e., ∃x[191(x) ∧ Pass(x)]
• Can show C if can find (remember: ∃ is a search) a 191 student who passes
• We know about one individual: Fred
• ∴ try to show (191(fred) ∧ Pass(fred)):

• to do that, need to show 191(fred) and need to show Pass(fred)

1. To show 191(fred) is easy: It's P3
2. To show Pass(fred):
use UI (x := fred) on P2, then MP if can show Frosh(fred)
Can show Frosh(fred) from P1 by UI (x := fred)
then MP if can show 191(fred)
But 191(fred) is easy: It's P3 (again)

5. Proof of validity:

 1 ∀x[191(x) → Frosh(x)] : P1 2 (191(fred) → Frosh(fred)) : 1; UI (x := fred) 3 191(fred) : P3 4 Frosh(fred) : 2,3: MP 5 ∀x[Frosh(x) → Pass(x)] : P2 6 (Frosh(fred) → Pass(fred)) : 5; UI (x := fred) 7 Pass(fred) : 4,6; MP 8 (191(fred) ∧ Pass(fred)) : 3,7; Conj 9 ∃x[191(x) ∧ Pass(x)] : 8, EG (fred := x)

3. Understanding & Creating Proofs

• (click on link & read the .ppt, .pdf, or .html file)
• (you might also find the article by Bowling (1977) of interest)
• (we'll look at "Proof Strategies" next time)