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I am reading through "Mathematical Logic by Ian Chiswell & Wilfred Hodges"(amazon, and publisher)

So far have it has covered $\land$-Introduction and $\land$-Elimination

Sadly this text only has answers to selected solutions, which annoys me to no end.


Exercise 2.3.3 (p.15) is Show that $\{\phi_1, \phi_2\} \vdash \psi$ if and only if $\{(\phi_1 \land \phi_2)\} \vdash \psi$

I am a little confused as to what to show here.


I can state $\{\phi_1, \phi_2\} \vdash (\phi_1 \land \phi_2)$ as I can build this using $\land$-Introduction, but I don't follow how this involves if-and-only-if.

I think this is in part because I am a little confused as to what "$\{\phi_1, \phi_2\}$" means - can this be read as "the set of assumptions we take to be true contains $\phi_1$ and $\phi_2$" ? - or maybe "the set of assumptions we take to be true is exactly $\phi_1$ and $\phi_2$" ?


edit: for context I am reading through this for self-study, so I don't have the normal support of a classroom environment - and the lack of exercise solutions makes it hard to check my understanding.

cjh
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2 Answers2

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I don't follow how this involves if-and-only-if.

Hint

A) For: if $\{ (φ_1 ∧ φ_2) \} \vdash ψ$, then $\{ φ_1, φ_2 \} \vdash ψ$.

1) $φ_1$ --- assumed

2) $φ_2$ --- assumed

3) $(φ_1 ∧ φ_2)$ --- from 1) and 2) by (∧I)

So far we have:

$\{ φ_1, φ_2 \} \vdash (φ_1 ∧ φ_2)$.

Thus, by Sequent Rule (Transitive Rule) [page 8], from it and $\{ (φ_1 ∧ φ_2) \} \vdash ψ$ we conclude with:

$\{ φ_1, φ_2 \} \vdash ψ$.

The Sequent Rule (Transitive Rule) says:

If $(Δ \vdash ψ)$ is correct and for every $δ \in Δ$, $(Γ \vdash δ)$ is correct, then $(Γ \vdash ψ)$ is correct, where $Δ$ and $Γ$ are sets of formulae.

In your example, we have:

$Γ = \{ φ_1, φ_2 \}$ and $Δ = (φ_1 ∧ φ_2)$.

The derivation 1)-3) is $Γ \vdash δ$, where $(φ_1 ∧ φ_2)$ is the unique $δ \in Δ$, and the premise: $\{ (φ_1 ∧ φ_2) \} \vdash ψ$ is $(Δ \vdash ψ)$.

Thus, by transitivity, we conclude with: $(Γ \vdash ψ)$.

B) For: if $\{ φ_1, φ_2 \} \vdash ψ$, then $\{ (φ_1 ∧ φ_2) \} \vdash ψ$, we have to use (∧E).

Mauro ALLEGRANZA
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  • Thank you very much for taking the time to work through this example, I've accepted the first answer - but I found both of them more helpful together than either on their own - so I had to flip a coin for the actual acceptance and the other answer (once clarified) helped me understand your answer. But again, thank you very much as without this I would have been lost - I really appreciate the detail you've included. – cjh Jun 27 '16 at 15:02
  • After coming back to this, your answer is self contained and fully worked, so I have accepted this. – cjh Jul 18 '16 at 05:59
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I think this is in part because I am a little confused as to what "{ϕ1,ϕ2}" means - can this be read as "the set of assumptions we take to be true contains ϕ1 and ϕ2" ? - or maybe "the set of assumptions we take to be true is exactly ϕ1 and ϕ2" ?

Indeed I think that's the only problem you're having here.

We write "$S \vdash φ$" to mean that we can derive the formula $φ$ by a sequence of rules from the formulae in the set $S$.

So for your exercise, the forward implication follows directly from $\land$-elimination, while the backward implication follows from $\land$-introduction.

If you need worked examples, you might want to try Stephen Simpson's notes.

user21820
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  • Your explanation of the definition of sequents and entails is the clearest I've found. Can you please clarify which direction of "implication" is forward? Left to right? As that would be if ${\phi1,\phi2}$ then ${(\phi1 \land \phi2)}$ - which to me seems to be using $\land$-introduction – cjh Jun 27 '16 at 03:30
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    @cjh: Yes my "forward" means "left-to-right"; sorry if that was not clear. But you're wrong in the second part of your comment. If you want to get from "A derives X" to "B derives X" you would want "B derives A"! This is because if A derives X and B derives A then B derives X. – user21820 Jun 27 '16 at 10:19
  • Thanks for that explanation, so I'm confused about forward and backward implication. Are you able to give or link me to concise definitions for forward and backward implication? Thank you very much for your time. – cjh Jun 27 '16 at 13:37
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    @cjh: It's not about definitions. It's about logical reasoning. The forward implication is of the form "if ( $A \vdash X$ ) then ( $B \vdash X$ )". To prove it, you have to show that in any situation where "$A \vdash X$" is true, "$B \vdash X$" is also true. How? In such a situation all you know is that there is a proof of $X$ from $A$, so you must figure out how to make use of such a proof. Clearly then you need to somehow from $B$ prove $A$. – user21820 Jun 27 '16 at 13:48
  • Again, thank you very much for your patience. I was initially thinking that for "if ($A \vdash X$) then ($B \vdash X$)" would require showing ($A \vdash B$) - but I now realise that is insufficient . if B is a subset of A then ($A \vdash B$) is true, however the set difference between A and B could contain the parts needed to prove X - therefore the only way to show this is to instead show that ($B \vdash A$) and therefore ($B \vdash X$) via transitive rule. Is that reasoning correct? – cjh Jun 27 '16 at 14:46
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    @cjh: Yes, that is correct. In general always think carefully through the logical meaning of statements, then you won't make this kind of error. (Unless it's a careless mistake, in which case I can't help you; I'm often quite careless!) – user21820 Jun 27 '16 at 14:50
  • Just one last clarification - here forward implication means to move forward from "if (A entails X) then (B entails X)" by showing (B entails A)?. Is backward implication moving from "if (B entails X) then (A entails X)" by showing (A entails B)? - thus completing both halves of the IFF? – cjh Jun 27 '16 at 14:53
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    @cjh: Sorry that's not what I meant. When we talk about an equivalence statement of the form "A if and only if B", the forward implication refers to the statement "If A then B" and the backward implication refers to "If B then A". Clearly we need to prove both implications in order to prove the original. There is no name given to the 'method' of proving used here. But in general note that such kinds of statements of the form "A gives B" have this behaviour if "gives" is transitive. Specifically if we want to prove "( A gives C ) implies ( B gives C )" it is enough to prove "B gives A". – user21820 Jun 27 '16 at 14:56
  • Ahh okay that makes more sense, here we just happen to be using transitivity - this confused me around the nomenclature as it felt like it was backwards - now forward and backwards make more sense. Thank you very much for all your time and patience :) – cjh Jun 27 '16 at 14:58
  • @cjh: Yup. You're welcome! You can accept my answer if it satisfies you. =) – user21820 Jun 27 '16 at 14:59