# What are the fundamental symbols and concepts in logic?

1. Jan 13, 2008

### ice109

what are the fundamental symbols in logic? maybe this is a vague question but that's cause i don't know anything about logic.

are they $$\neg$$ $$\wedge$$ $$\vee$$?

do we define the material implication in terms of these symbols?

basically i'm having trouble understanding the paradoxical nature of the material implication and so i'm wondering what it's supposed to accomplish? by this i mean how was it decided for which values it is true. please illustrative examples because they don't suffice to explain to me why the truth table for -> is what it is.

2. Jan 13, 2008

### Hurkyl

Staff Emeritus
(note: this post is trying to be at an intuitive, motivational level, and not a technical level)

(note: In this post, => will be the syntactic symbol, and -> will denote the binary function of truth values)

One important rule of inference in classical logic is:

P
P => Q
Q

We also have a few things that are not rules of inference:

P => Q
Q

Q
P => Q
P

If we use a two-valued system of truth values, can you derive what I just said from the truth table for ->? What if the truth table was different?

Last edited: Jan 13, 2008
3. Jan 13, 2008

### ice109

i appreciate your tailoring the explanation. i don't quite understand what you mean by derive from the truth table?

maybe you mean find the corresponding line in the truth table?

4. Jan 13, 2008

### Hurkyl

Staff Emeritus
I'll demonstrate for modus tollens:

~Q
P => Q
~P

Now, let v be a binary truth valuation -- a function that assigns binary truth values to propositions, and has the right relationship with the logical connectives. There are only four ways it can assign truth values to the propositions Q, P, and P => Q:

v(P) = T, v(Q) = T, v(P => Q) = T
v(P) = T, v(Q) = F, v(P => Q) = F
v(P) = F, v(Q) = T, v(P => Q) = T
v(P) = F, v(Q) = F, v(P => Q) = T

(Because v(P => Q) must be the same as v(P) -> v(Q))

Now, suppose ~Q and P => Q are valid statements with respect to v; that is, v(~Q) = T and v(P => Q) = T. Consulting the four possibilities, we see only one remains:

v(P) = F, v(Q) = F, v(P => Q) = T

and so we have the following calculation for truth values:
If we have
1. v(~Q) = T
2. v(P => Q) = T
Then we also have
3. v(~P) = T​

So we see that from the truth table for ->, we have derived a rule for truth values analogous to modus tollens.

Last edited: Jan 13, 2008
5. Jan 13, 2008

### ice109

ok i understand

for modus ponens ( i think that's what it is )

v(P)=T, v(P=>Q)=T our only option from our truth table for v(Q)=T

and i could do the others. my question is => defined by modus ponens or somehow else?

6. Jan 13, 2008

### Hurkyl

Staff Emeritus
Just to make sure it's clear, I think your question is purely one of exposition, and it's going to depend upon what you mean by "defined by".

Syntax essentially has only three things:
(1) An alphabet of symbols
(2) A grammar that tells you when an arrangement of symbols forms a predicate
(3) A specification of which rules of inference are admissible

(A rule of inference is a means to take a collection of predicates and compute a new predicate. e.g. modus ponens is the rule that produces Q given {P, P => Q})

So, syntactically, => is just a symbol of the alphabet, and there really isn't anything more to it.

7. Jan 13, 2008

### ice109

my confusion is that this symbol connotes something in argument. i want to be able to apply what i know about the word implication in interpreting this symbol. am i seriously supposed to just memorize the truth table?

Last edited: Jan 13, 2008
8. Jan 13, 2008

### Hurkyl

Staff Emeritus
(I'm continuing to use => and -> as in post #2)

The notion of "deductive argument" is separate from the notion of "truth".

=> is just a symbol of the alphabet, and we have rules of inference that tell us how to infer new predicates from old ones. I suppose you could say that the totality of such rules (e.g. modus ponens, modus tollens, etc) 'defines' the role of => in logical inference.

-> is just an operation for manipulating truth values. It doesn't directly have any relation to the notion of argument. However...

It is a fact of classical logic that the following three things are equivalent:

(1) This is a rule of inference of classical logic:
P1, P2, ..., Pn
___________
Q

(2) This is a tautology in classical logic:
$(P_1 \wedge P_2 \wedge \cdots \wedge P_n) \implies Q$

(3) For every Boolean truth valuation v:
$\left( v(P_1 \wedge P_2 \wedge \cdots \wedge P_n) \rightarrow v(Q) \right) = \top$

(It takes quite a lot of effort to demonstrate the equivalence, however)