Is $\large \frac{\varphi}{\psi\to \varphi}$ a correct derivation using just $(\to I)$?

Question

In Dirk Van Dalen's text "Logic and Structure" third edition (page 33) the following line is written:

Furthermore one may apply $(\to I)$ if there is no hypothesis available for cancellation e.g.$ \large \frac{\varphi}{\psi\to \varphi} $ is a correct derivation using just $(\to I)$

I'm a little confused with this;I can't see how its a correct derivation according to the (usual) inductive definition of a derivation given in def 1.4.1. I wonder if this is a typo for $ \large \frac{\varphi}{\varphi\to \varphi}$ which I think is a correct derivation [with the hypothesis chosen to remain uncancelled]

No, it is correct. You can consider that $\varphi \to (\psi \to \varphi)$ is a tautology. — Mauro ALLEGRANZA, Oct 26 '21 at 13:15
The issue is: we can always add "unnecessary" assumptions; see Weakening. See Lemma 2.4.3. (a) page 35. — Mauro ALLEGRANZA, Oct 26 '21 at 13:17
@MauroALLEGRANZA But how to prove it from the inductive definition? — Vivaan Daga, Oct 26 '21 at 13:18
In a nutshell we have: $\varphi \vdash \varphi$ and thus $\psi, \varphi \vdash \varphi$, from which - by $(\to \text I)$ we have $\varphi \vdash \psi \to \varphi$. — Mauro ALLEGRANZA, Oct 26 '21 at 13:23
@MauroALLEGRANZA There is no such weakening rule given in the inductive definiition — Vivaan Daga, Oct 26 '21 at 13:25
"Definition 2.4.2 The relation $\Gamma \vdash \varphi$ between sets of propositions and propositions is defined as follows: there is a derivation with conclusion $\varphi$ and with all (uncanceled) hypotheses in $\Gamma$." Thus $\psi,\varphi \vdash \varphi$ is a derivation. — Mauro ALLEGRANZA, Oct 26 '21 at 13:28
The inductive def I'm talking about are given in page 35 here http://inis.jinr.ru/sl/vol2/Ax-books/Disk_01/Dalen-Logic-and-Structure.pdf — Vivaan Daga, Oct 27 '21 at 12:53
possible duplicate question. https://math.stackexchange.com/questions/4289794/error-in-logic-and-structure — nickalh, Oct 28 '21 at 10:56

score 2 · Accepted Answer · answered Oct 26 '21 at 13:55

2

$\dfrac{\varphi}{\psi\to\varphi}{\small({\to}\mathsf I})$ is valid. The rule of conditional introduction allows you to discharge as many of the assumptions of the antecedent ($\psi$) that have been made; including none of them. So it is okay that none have been assumed. You may safely discharge all of them; which is none.

Since $\varphi\vdash\varphi$ is valid, $\psi,\varphi\vdash\varphi$ is too, and so you may infer $\varphi\vdash\psi\to\varphi$.

answered Oct 26 '21 at 13:55

Graham Kemp

129,094

According to the book only if $\psi$ is an assumption you're allowed to discharge so how to make it an assumption using the inductive laws? – Vivaan Daga Oct 26 '21 at 15:45
The inductive def im asking about are given in page 35 here http://inis.jinr.ru/sl/vol2/Ax-books/Disk_01/Dalen-Logic-and-Structure.pdf – Vivaan Daga Oct 27 '21 at 12:55
https://math.stackexchange.com/questions/4289794/error-in-logic-and-structure?noredirect=1&lq=1 I summarised my confusion here – Vivaan Daga Oct 28 '21 at 10:59
Why isn't it $\dfrac{\varphi \space \space [\psi]}{\psi\to\varphi}$ – Vivaan Daga Oct 28 '21 at 14:54
Can you please state the implies intro rule formally? – Vivaan Daga Oct 28 '21 at 14:56
Are the rules https://staff.fnwi.uva.nl/b.vandenberg3/Onderwijs/Proof_Theory_2016/handout_2.pdf given here OK ? – Vivaan Daga Oct 28 '21 at 16:26
Why the downvote? Does my question really deserve to be downvoted? – Vivaan Daga Jul 30 '22 at 04:11

Is $\large \frac{\varphi}{\psi\to \varphi}$ a correct derivation using just $(\to I)$?

1 Answers1