Predicate Logic Flashcards

1
Q

Existential Quantifier

A

∃, “for some” or “there is” or “there exists”

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Universal Quantifier

A

∀, “every” or “all” or “each”

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

∀ and ∃ Equivalence

A

∃x is equivalent to ¬∀x¬

∀x is equivalent to ¬∃x¬

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Individual constants

A

a, b, c etc.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Individual variables

A

x, y, z etc.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Predicate letters

A

P, Q, R etc.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Function symbols

A

f, g, h etc.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Term

A

Expressions referring to individual objects, e.g. individual constants, individual variables, functions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Formula

A

Terms, terms with propositional connectives, formula

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Ground term

A

A term which contains no variables and is constructed entirely from constants and function symbols

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Bound

A

All occurrences of x in a quantified formula ∀xα or ∃xα are said to be bound

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Free

A

Any occurrence of a variable that is not bound is said to be free

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Closed formula

A

A formula containing no free variables is said to be closed

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Open formula

A

A formula containing at least one free variable is said to be open

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

And-introduction

A

From φ and ψ you can infer φ ∧ ψ

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

And-elimination

A

From φ ∧ ψ you can infer both φ and ψ

17
Q

Or-introduction

A

From φ you can infer both φ ∨ ψ and ψ ∨ φ

18
Q

Or-elimination

A

If you can infer χ both from φ and from ψ, then you can infer χ from φ ∨ ψ

19
Q

If-introduction

A

If you can infer ψ from φ, then you can prove φ → ψ

20
Q

If-elimination

A

From φ and φ → ψ you can infer ψ