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Give an equational proof of $ \vdash (\forall x)(A \rightarrow B) \equiv ((\exists x) A) \rightarrow B$

I tried and used ping pong theorem to split it into two implications to prove

$$ \vdash (\forall x)(A \rightarrow B) \rightarrow (((\exists x)A) \rightarrow B)$$

and

$$ \vdash (((\exists x)A) \rightarrow B) \rightarrow (\forall x)(A \rightarrow B) $$

But still couldn't figure it out.

Lists of axioms and theorems.

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Out Of Bounds
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2 Answers2

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(Like the other answer, I will assume that $\;x\;$ is not free in $\;B\;$. Otherwise this is not a theorem: take for example $\;A := x > 2\;$ and $\;B := x > 0\;$, which leads to a statement that is false for $\;x \le 0\;$.)

(I will use a slightly different quantification notation, to save on parentheses.)

If you only want to use the axioms and theorems from the photos you provided, then you seem to be missing something: there is no axiom about $\;\exists\;$. So I will assume you are allowed to use, e.g., the following definition the of $\;\exists\;$: $$ \tag{0} \langle \exists x :: A \rangle \;\equiv\; \lnot \langle \forall x :: \lnot A \rangle $$

Using $(0)$, the proof is simply as follows:

\begin{align} & \langle \forall x :: A \to B \rangle \\ \equiv & \qquad \text{"by (2.4.11) -- working towards (6.4.2),} \\ & \qquad \phantom{\text{"}}\text{which seems the key rule to use here"} \\ & \langle \forall x :: \lnot A \lor B \rangle \\ \equiv & \qquad \text{"by symmetry (6) -- still towards (6.4.2)"} \\ & \langle \forall x :: B \lor \lnot A \rangle \\ \equiv & \qquad \text{"by (6.4.2), using the assumption that $\;x\;$ is not free in $\;B\;$"} \\ & B \lor \langle \forall x :: \lnot A \rangle \\ \equiv & \qquad \text{"introduce double negation (2.4.4)} \\ & \qquad \phantom{\text{"}}\text{-- to match the RHS of (0): we work towards $\;\exists\;$"} \\ & B \lor \lnot \lnot \langle \forall x :: \lnot A \rangle \\ \equiv & \qquad \text{"by duality (0)"} \\ & B \lor \lnot \langle \exists x :: A \rangle \\ \equiv & \qquad \text{"by symmetry (6) -- to reintroduce $\;\to\;$"} \\ & \lnot \langle \exists x :: A \rangle \lor B \\ \equiv & \qquad \text{"reintroduce $\;\to\;$ by (2.4.11)"} \\ & \langle \exists x :: A \rangle \to B \\ \end{align}

which completes this proof.

(To make this formally complete, we must of course use symmetry $(2)$, Leibniz $(2.1.16)$ and $(6.1.11)$, and transitivity $(1.4.13(c))$ many times, but fortunately that is implicit in the above proof format designed by Edsger W. Dijkstra and Wim Feijen. See EWD1300 for more details.)

MarnixKlooster ReinstateMonica
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  • Probably you can make it a acceptable proof after adding many of the missing steps you refer to, but for a complete proof you should write out all those steps. (they are just an integral part of the proof. As it is, I would not even accept the first step , you seem to use a rule $\forall x (D) $ , $ D \equiv E $ then $\forall x (E) $. This inference is not allowed, as stated the only allowed rule is $ D $ , $ D \equiv E $ then $ E $ (so without the $ \forall x$ ), I do think that in a proof all steps should be explicit. – Willemien Apr 14 '14 at 06:28
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    @Willemien Acceptable to whom, in which context? Your standard that "in a proof all steps should be explicit" rules out all not-completely-formal proofs. I would argue that the proof from this answer is perfectly acceptable in the context of equational proofs in the style of Dijkstra-Feijen, Gries-Schneider, and probably Tourlakis, where they tries to teach how to use logic of just studying it, and then basically ignoring it when doing 'real' mathematics. – MarnixKlooster ReinstateMonica Apr 14 '14 at 15:45
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    @Willemien The first two steps use Leibniz in the form $(6.1.11)$ from the original images. I've made that explicit now in the remark at the end of my answer. – MarnixKlooster ReinstateMonica Apr 14 '14 at 15:46
  • I think you should produce a formal proof when asked to do so, no hand waving by remarking you can make it complete by using this theorem or derrived rule. No, just give the complete proof, even your remark that "The first two steps use Leibniz in the form (6.1.11)" is I think wrong, Does "dnof" not mean "does not appear free" and $x$ does appear free in $A$ so you cannot use this rule. (even forgetting that the OP only gave From $ A $ and $ A \equiv B $ you may conclude $B$ as inference rule. Also I object to introducing theorems in the proof, only axioms are allowed to be introduced. – Willemien Apr 14 '14 at 22:31
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    @Willemien The OP asked for "an equational proof", not a completely-formal-not-even-using-theorems proof. And when glancing through the book, Mathematical Logic by George Tourlakis, and e.g., looking at page 65, I now see that that my proof closely matches what Tourlakis calls an equational proof. – MarnixKlooster ReinstateMonica Apr 15 '14 at 04:23
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    @Willemien According to its side condition, $(6.1.11)$ can be used if $;x;$ does not appear free in $;\Gamma;$: it is allowed for $;x;$ to occur freely in $;A;$. So as far as I can see I'm justified to use that rule. – MarnixKlooster ReinstateMonica Apr 15 '14 at 04:26
  • Are we discussing here a line in a proof that has not even been written down? What more do you want as proof that the proof is incomplete? and is it not a goal of logic to make proofs without holes. even when the inference is valid, (what it isn't because it is using an inference rule that the OP didn't mention), the line should be present, but in this case it isn't there. – Willemien Apr 15 '14 at 07:50
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    We're discussing the first two steps in my proof, each of which requires an additional to formally justify; for example, the first step is justified by $(2.4.1)$ together with $(6.1.11)$, and (in line with the OP's book) I made a conscious choice to leave the latter implicit in the calculation, since this Leibniz-like rule is part of the basic mechanics of any equational proof. (I'm ending this discussion for my part, since it stopped being productive; it looks like we just differ on what is "acceptable" in this context.) – MarnixKlooster ReinstateMonica Apr 15 '14 at 17:32
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It is all to horrible, dare your lecturer to prove it himself.

Equational proofs are more complex than Hilbert style (modus ponens only proofs)

Hilbert style (modus ponens only proofs) are more complex than the normal natural deduction proofs.

And in Natural deduction the proof goes as follows:

I presume that there is no x free in B and for sanity's sake I will use Ax to indicate that A can contain x.

At is just shorthand for A(x:=t) , Au for A(x:=u), Vx stands for $\forall x$ and Ex for $\exists x$

1. | |______ (Vx)(Ax -> B)                         Assumption for conditional proof
2. | | |____ (Ex)AX                                Assumption for conditional proof 
3. | | | |_t At                                    Assumption for existentional elimination proof
4. | | | |   At -> B                               1 Universal elimination
5. | | | |   B                                     3,4 modus ponens
.. | | | <------------------------------------------ end of sub proof
6. | | |     B                                     2,3-5  existentional elimination
.. | | <------------------------------------------- end of sub proof
7. | |       ((Ex)AX) -> B                         2-6 conditional proof 
.. | <---------------------------------------------- end of sub proof 
8. |         ((Vx)(Ax -> B)) ->  ((Ex)AX) -> B )   1-7 conditional proof  
9. | |______ ((Ex)AX) -> B )                       Assumption for conditional proof
10 | | |___u                                       variable for universal introduction
11 | | | |__ Au                                    Assumption conditional proof
12 | | | |   (Ex)AX                                11 existentional introduction
13 | | | |   B                                     9,11 modus ponens
.. | | | <------------------------------------------ end of sub proof
14 | | |     At -> B                               11-13 conditional proof
.. | | <-------------------------------------------- end of sub proof
15 | |       (Vx)(Ax -> B)                         10-14 Universal introduction
.. | <---------------------------------------------- end of sub proof 
16 |         ((Ex)AX) -> B ) -> ((Vx)(Ax -> B))    9-15 conditional proof
17 |         ((Vx)(Ax -> B)) <->  ((Ex)AX) -> B )  8,16 Equivalence introduction

But this is an natural deduction proof.

And you are asked to transform it in an equational proof?

I could tell you to first transform the proof above into an Hilbert style (modus ponens only) proof and then transform that in an equational proof.

but that would even be to hard for me so to be honnest I don't give you a 5% chance to succeed.

Just dare your lecturer to do it.

If he manages it, publish the proof here, (then I can also learn from it) and if he fails, (what I would not find improbable) he should never have asked you.

All the best.

Willemien
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    I don't really understand your attitude towards equational proofs. I just added an equational proof in a separate answer. Assuming it is correct, how is that proof horrible, or more complex than your natural deduction proof? – MarnixKlooster ReinstateMonica Apr 14 '14 at 04:58