# A (predicate) logical expression and its meaning

I will use an example sentence from Russell, so if we let W represent the predicate “... authored Waverley “, then:

(x)(Wx → (y)(Wy → y=x))

would express the same thing as the sentence "At most one person authored Waverley".

Now if one would introduce (z) would it matter? If one put in the following expression:

(z)(Wz → z=x)

would it change the meaning, or would it only be redundant? Would the following:

(x)(Wx → (y)(z)(Wy → y=x )∧(Wz → z=x))

then be the same as

(x)(Wx → (y)(Wy → y=x)) ?

• There is a connective (like "and") inside : **(Wy → y=x )(Wz → z=x) ??? If not, the formula is "ill-formed" i.e. not correct according to the syntactical rules of the language. – Mauro ALLEGRANZA Oct 8 '14 at 12:58
• Thanks, so if I add '∧' to (Wy → y=x )(Wz → z=x) like this: (Wy → y=x )∧(Wz → z=x) it would be syntactically correct? @Mauro ALLEGRANZA – WaWaWa Oct 8 '14 at 13:07
• Yes, now it is... and we can say that it is superfluous. "P and P" is equivalent to "P"; thus (y)(z)[(Wy → y=x )∧(Wz → z=x)] add nothing to (z)(Wz → z=x) – Mauro ALLEGRANZA Oct 8 '14 at 13:19
• Thanks,@MauroALLEGRANZA as I suspected. So if one would have substituted the 'z' with a 'y', the identity (if one can use that word here) would have been easier to recognise, because one would simply have used the expression (y)(Wy → y=x) twice? However 'z' and 'y' as signs are obviously different, but that don't matter at all? Am I correct? So z = y in this case? – WaWaWa Oct 8 '14 at 14:00
• In a quantified expression like (z)(Wz → z=x) the bound variable z in not meaningful, i.e. we can change it with any other (like y) without affecting the meaning, provided that we do not use a variable already present in the formula and free (like x in the example), because in this case we have problems. – Mauro ALLEGRANZA Oct 8 '14 at 14:03

## 1 Answer

The z would be completely redundant. This is because 'what you're doing with z', i.e. (Wz → z=x), is the same as what you're doing with y, and they are both from the same set; that is, every z could've been y and vice versa. This together means every case of z is already caught by y.

### Longer explanation:

There is a very intuitive mathematical rule that says that we can change the name of bound variables (variables can be bound by existential or universal quantification). Basically, it means that the following two are equivalent:

• (x)(P(x))
• (y)(P(y))

'x' or 'y' are just names, and they don't carry meaning.

There is another rule that says we can swap universal quantifications as we like, such that the following two are equivalent:

• (x)(y)(P(x) ∧ Q(y))
• (y)(x)(P(x) ∧ Q(y))

Note that this is not the same rule as above, i.e. we didn't rename x to y and vice versa, because the inner experssion "P(x) ∧ Q(y)" doesn't change. Only the order of the quantifications is changed.

Intuitively, these two together show that we may rename the z you added to y, yielding:

(x)(Wx → (y)(y)(Wy → y=x) ∧ (Wy → y=x))

Note: this is no good notation, because the y's are ambiguous. This is just to give you the idea of the intuitive notion.

On both sides of the ∧ we have the same expression, so we can simplify it:

(x)(Wx → (y)(y)(Wy → y=x))

And then we can also remove one of the y's...

(x)(Wx → (y)(Wy → y=x))

Bringing us back to the initial expression.