## BRACERS Record Detail

To access the original letter, email the Ready Division.

Churchill College Archives Centre, Cambridge

Newer copies of his material from the Churchill Archives Centre are also available as part of Rec. Acq. 1816, Box 16.79, part A.

Cambridge University, Chuchill College, Churchill Archives Centre; Ralph Hawtrey Archvies, HTRY 10/81A.

**BR TO RALPH G. HAWTREY,** **22 JAN. 1907**

BRACERS 55816. ALS(X). Churchill College, Cambridge. *Russell* 22 (2002): 158–9, 160; **4e **in *Papers* 5

Edited by B. Linsky and K. Blackwell. Reviewed by N. Griffin

Bagley Wood.^{1}

Jan. 22. ’07

Dear Hawtrey,^{2}

I forgot to send you the paradox which pilled^{3} the substitution-theory. Here it is. Put

$p_{\scriptscriptstyle 0} \, .= \: :(∃ \, p, a):a_{\scriptscriptstyle 0} \, .= \: .p \frac{b}{a} ! \, q: \: ∼ (p \frac{a_0}{a})$ [where “$p \frac{b}{a} ! \, q$” means “$p$ becomes $q$ by substituting $b$ for $a$”]

^{4} Then

$$p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle <0>}} ! \, q}{a_{\scriptscriptstyle 0}} \, .= \: :(∃\,p, a): p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle 0}} ! \, q \, .= \:.\, p \frac{b}{a}! \, q \, : \; \sim \Bigg(p \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a} \Bigg)$$ ^{5} Hence

$$p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle <0>}} ! \, q}{a_{\scriptscriptstyle 0}} \, . \supset \: : (∃\,p, a): p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle 0}}! \, q \, .= \:.\, p \frac{b}{a}! \, q: p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a_{\scriptscriptstyle 0}} \, . \; ∼ \Bigg(p \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a} \Bigg) \tag 1$$

$$∼ \Bigg(p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a_{\scriptscriptstyle 0}} \Bigg) \,. \supset \: :.\, p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle 0}}! \, q \, .= \:.\, p \frac{b}{a}! \, q\, : \: \operatorname*{\supset}_{p, a} \, .\, p \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a}:. \supset \: :.\, p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a_{\scriptscriptstyle 0}} \tag 2$$

^{a} ^{, } ^{b}

$$(1) \, .(2) \, . \supset \: :.\, p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a_{\scriptscriptstyle 0}} :.(∃ \, p, a):p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle 0}} ! \, q \, .= \:.\, p \frac{b}{a}! \, q \, : \; ∼ \Bigg(p \frac{p_{\scriptscriptstyle 0} \frac{b}{a_0} ! \, q}{a} \Bigg) \tag 3$$But if $p_{\scriptscriptstyle 0} \frac{b}{a_0}! \, q$ is the same prop^{c} as $p \frac{b}{a}! \, q$, it seems plain we must have $p=p_{\scriptscriptstyle 0} \, .a=a_{\scriptscriptstyle 0}$, whence $p \frac{p_0 \frac{b}{a_0} ! \, q}{a} \, . ≡ \:.\, p_{\scriptscriptstyle 0} \frac{p_0 \frac{b}{a_0} ! \, q}{a_0}.$ Thus it is impossible that $p \frac{p_0 \frac{b}{a_0} ! \, q}{a}$ should be false while $p_{\scriptscriptstyle 0} \frac{p_0 \frac{b}{a_0} ! \, q}{a_0}$ is true, which, by (3), is shown to be involved.

In trying to avoid this paradox, I modified the substitution-theory in various ways, but the paradox always reappeared in more and more complicated forms.

Yrs ever

B. Russell.

Typeset by A. Duncan 18/01/2019

### Notes

- 1.
**[document]**The letter was edited from a photocopy given to the RA about 1980, when the original was in private hands. Now it is in the Hawtrey papers in the Churchill Archives Centre, Churchill College, Cambridge, who provided a photograph. It was previously published in Bernard Linsky, “The Substitutional Paradox in Russell’s 1907 Letter to Hawtrey [corrected reprint]”,*Russell*22 (2002): 151–60 (with facsimile on 160); Linsky proposes two corrections to Russell’s text (see n. 5). Also published as**4e**in*Papers*5. - 2.
**Hawtrey**Ralph George Hawtrey (1879–1975) was an economist with the UK Treasury from 1904. At Trinity College, Cambridge, where he was an Apostle, Hawtrey took a 1st class in the mathematics tripos in 1901. - 3.
**forgot to send you the paradox which pilled**Hawtrey and BR must have been in communication, but no previous correspondence or meetings between them are known; next year they had several exchanges on logic. To*pill*is to “fail (a candidate) in an examination” (*OED*, 3rd ed., where the word has no quoted uses before 1908 [cited by Linsky, p. 157 n. 12]; or to “reject or exclude by ballot (a person) from membership of a club or society; to blackball”). For more on uses of “pilled”, see*Papers*5, A125: 2–3. - 4.
**Then $p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle <0>}} ! \, q}{a_{\scriptscriptstyle 0}}$****… $\sim \Bigg(p \frac{p_0 \frac{b}{a_0} ! \, q}{a} \Bigg)$**In the first term of this equation, Russell carelessly omitted the subscript “$0$” to the “$a$” in “$p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle <0>}} ! \, q$”. He made the same mistake in the following line as well, presumably because he was simply copying what he had written in the line above. In both cases the omission has been editorially corrected by adding the missing symbol in angle brackets. The definition of “$p_0$” given in the first line of the proof makes it clear that “$p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle 0}} ! \, q$”, rather than “$p_{\scriptscriptstyle 0} \frac{b}{a} ! \, q$”, has to be substituted for “$a_0$” in order to get the term which immediately follows the quantifiers in the present line.As the image of the letter reveals, in the final term of this equation the subscript “$0$” to the “$a$” in “$p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle 0}} ! \, q$” has been circled and a question mark placed alongside the line. This was not done by Russell and has not been included in the transcription. Presumably it was done by Hawtrey, who no doubt noticed that the substitutions Russell made in this line were inconsistent — for the substitution made for “$a$” in the last term must be the same as that made for “$a_0$” in the first — but mistakenly thought that it was the last term that was in error.

The issue is described in more detail in Linsky, where a detailed reconstruction of the argument is also given. The text is transcribed uncorrected in

*Papers*5. - 5.
**Hence $p_{\scriptscriptstyle 0} \frac{p_{\scriptscriptstyle 0} \frac{b}{a_{\scriptscriptstyle <0>}} ! \, q}{a_{\scriptscriptstyle 0}}$**See note 4 above.

### Textual Notes

- a.
**$\boldsymbol{(1).}$**Following a deleted “ $\vdash$ .” - b.
**$\boldsymbol{(2). ⊃}$**Followed by a cancelled “ $\vdash \: :: \text {Hp} \: . \: \supset$”. Linsky comments: “This stylistic change, while not altering the logical force of the line, does make it more clearly a proof from hypotheses.” - c.
**prop**BR’s logical slang for “proposition”.

**Record created**1994/05/09

**Record last modified**2023/10/25

**Created/last modified by**duncana