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My teacher has assigned us this exercise as part of our homework:

Give a natural deduction proof of $(\alpha\to\beta)\to(\beta\to\gamma)\to(\alpha\to\gamma)$

Here is an example of natural deduction that he gave us:

$\vdash \alpha\to\beta\to\alpha$.

This is a tautology, but it's easier to prove than to verify:
$\alpha,\beta\vdash\alpha$, hypothesis
$\alpha\vdash\beta\to\alpha$, deduction theorem
$\vdash\alpha\to\beta\to\alpha$, deduction theorem

Here is my attempt:

Hypothesis: $(\alpha\to\beta),(\beta\to\gamma) \vdash (\alpha\to\gamma)$
Deduction thm: $(\alpha\to\beta) \vdash(\beta\to\gamma)\to(\alpha\to\gamma)$
Deduction thm: $\vdash(\alpha\to\beta)\to(\beta\to\gamma)\to(\alpha\to\gamma)$

Is this enough? I feel like I need to take apart the functions inside the parentheses and check those as well, but I don't know how.

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  • $\begingroup$ Is your "hypothesis" an axiom of your natural deduction system? If so, your proof is fine. If not, you need to derive it from whatever your axioms are. $\endgroup$ – Lord_Farin Nov 19 '13 at 17:46
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Unless you already have that hypothesis, you need to get to that hypothesis first.

Also, (α→β)→(β→γ)→(α→γ) is ambiguous. You almost surely want to prove ((α→β)→((β→γ)→(α→γ))). Additionally, you don't actually use the deduction (meta) theorem, you use the rule of inference that it implies as valid. I'll call this rule "conditional introduction".

I don't know your format exactly or what you have exactly, but a proof might go something like this:

Hypothesis (rule of modus ponens/detachment): (α→β), α $\vdash$ β

Weakening: (α→β), α, (β→γ) $\vdash$ β

Commutation on the left side: (α→β), (β→γ), α $\vdash$ β

Identity: (α→β), (β→γ), α $\vdash$ β, (β→γ)

Detachment: (α→β), (β→γ), α $\vdash$ γ

Conditional introduction: (α→β), (β→γ) $\vdash$ (α→γ)

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  • $\begingroup$ Thank you! Everything makes sense to me in this proof, although I have one question. Is "Weakening" a deductive step in the proof, or just a necessary assumption in order to complete the proof? $\endgroup$ – Jeff Nov 20 '13 at 1:00
  • $\begingroup$ @Jeff I'm not sure. We'd need to know the specifics of your formal system. With a formal system we need to know that every step of a given proof satisfies the definition of a proof. This can mean that every step is either an assumption or hypothesis, an axiom, or follows from previous steps in the proof solely by the rules of inference permitted. $\endgroup$ – Doug Spoonwood Nov 20 '13 at 4:14
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$$\begin{align} (1) & \alpha \rightarrow \beta && [\text{HYP}] \\ (2) & \beta \rightarrow \gamma && [\text{HYP}] \\ (3) & \alpha && [\text{HYP}] \\ (4) & \beta && [\text{MP}(1,3)] \\ (5) & \gamma && [\text{MP}(2,4)] \\ (6) & \alpha \rightarrow \gamma && [\rightarrow\text{-intro}(3,5)] \\ (7) & (\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma) && [\rightarrow\text{-intro}(2,6)] \\ (8) & (\alpha \rightarrow \beta) \rightarrow ((\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma)) && [\rightarrow\text{-intro}(1,7)] \\ \end{align}$$

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  • $\begingroup$ Not much left to do here. $\endgroup$ – Dan Christensen Nov 20 '13 at 0:35
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If, as Doug suggested, you are really required to prove

$A\implies B \implies (B\implies C \implies (A\implies C))$,

you could start with 3 successive premises, the first two being:

  1. Suppose $A\implies B$

  2. Suppose $B\implies C$

  3. Suppose ???

(I don't like Greek letters!)

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I will replace $\alpha$ with $A$, $\beta$ with $B$, and $\gamma$ with $C$ to use the truth table generator and Fitch-style natural deduction proof checker.

To resolve the question of ambiguity, the following truth table shows that one cannot prove $((A\to B)\to(B\to C))\to(A\to C)$:

enter image description here

The counterexample is in the fourth line of the table where $A=T$, $B=F$, and $C=F$.

With the ambiguity resolved, here is a natural deduction proof:

enter image description here


Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

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