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I am looking for feedback to three proofs (alternatively derivations) that I have constructed. The first is:

Theorem. Injectivity does not imply surjectivity.

Proof: Suppose $\{\phi\} \vdash \theta$. Then according to the soundness theorem we have $\{\phi\} \models \theta$, which is to say that for every valuation $v$, $[[\phi]]_{v}=1 \implies [[\theta]]_{v}=1$. However, consider the following structure $\langle \mathbb{N};; +, \cdot, 1 \rangle$ which gives rise to the following formula $\exists x_{1}(x_{0} \dot{=} f_{2}(f_{1}(\overline{c}_{0},\overline{c}_{0}),x_{1}))$.Now, from this formula we define a function as follows:

$$f:\mathbb{N} \to \mathbb{N}, $$ $$f(x)=2x$$

This function is clearly injective. However it is equally clear that it is not surjective, as there are elements in the range that the function does not map to, more specifically all odd numbers. Thus, we have a case where $[[\{\phi\}]]_{v}=1$ while $[[\{\theta\}]]_{v}=0$, which shows $\{\theta \}$ is not a logical consequence of $\{\phi\}$ ($\{\phi\} \not \models \{\theta\}$). Hence, by the completeness theorem we have $\{\phi\} \not \vdash \theta$, or in more informal terms, the injective property does not imply the surjective property. $\square$

The two other proofs are a bit more complicated to present. As far as I can tell (and believe me I have tried), Math Stack Exchange does not support bussproof, which is a LateX package that enables users to construct proof trees. Furthermore, I haven't been able to simply attach a PDF file of the derivations (again Math Stack Exchange does not support it). As a result, I have decided to post my codes, hoping you (the community) still are willing to provide some feedback.

enter image description here

Community
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user100561
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3 Answers3

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  1. The counterexample to injectivity implying surjectivity is presented in a quite unnecessary laborious way. You will be marked down for obfuscation.

  2. The universally quantified wffs at the top of both branches of the first displayed proof shouldn't be at the same level as the existential. Move the existentials down five lines, and then you need to repeat the conclusion of the sub proof. Thus ...

$\quad\quad\quad\quad\quad\quad\varphi\\ \quad\quad\quad\quad\quad\quad\ldots\\ \quad\quad\quad\quad\quad\quad\ldots\\ \quad\quad\exists x\varphi\quad\quad C\\ \quad\quad -----\\ \quad\quad\quad\quad C$

See van Dalen section 2.9, or other texts.

Here's the proof done Fitch-style: Natural Deduction and Identity .

[Yes, it would be good if we could use bussproofs here!]

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Peter Smith
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  • But, don't I need to have them there in order to use the rule $\exists E$? – user100561 Oct 13 '13 at 14:02
  • They are needed but are misplaced. They should appear higher up so the conclusion of the sub proof and the existential you are going to discharge are at the same level. – Peter Smith Oct 13 '13 at 17:31
  • That is interesting, I thought I could resurrect my derivation by simply moving the hypothesis $\forall x_{1} a \dot{=} x_{1}$ one step upwards, then use $\forall E$ to obtain $a \dot{=} b$. But I take your answer as a no. Right? – user100561 Oct 13 '13 at 18:12
  • Moreover, since you did not voice any comments regarding my second derivation, I take it as a sign that it is valid. – user100561 Oct 13 '13 at 19:06
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I have made changes to the derivations - does this look better? Moreover, is the second derivation convincing (or rightly performed)?

enter image description here

user100561
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  • No, in 0.2, $\to E$ is evidently not a valid modus ponens. – Peter Smith Oct 15 '13 at 12:36
  • How should I address that? Can I remove the implication part of $f_{1}(x) \dot{=} f_{1}(y) \to x \dot{=} y$ and use the $\text{ers}$-rule for substitution, and thus shift the $\to E$ one step down? Evidently, I would have to derive $f_{1}(x) \dot{=} f_{1}(y) \to x \dot{=} y$ again from the injective hypothesis. – user100561 Oct 15 '13 at 13:39
  • After all $f(f(x)) \dot{=} f(x) \implies f(x) \dot{=} x$. – user100561 Oct 15 '13 at 13:46
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I have a new suggestion as to how to correct my mistake in 0.2; namely I introduce an assumption $f(f(x)) \dot{=} f(a)$ as the minor premise while I at the same time derive $f(f(x)) \dot{=} f(a) \to f(x) \dot{=}a$ as the major premise (from injectivity). Then, I use $\exists E$ at the next step to conclude $f(x) \dot{=} a$. From thereon, I use the Idempotence property to arrive at my desired conclusion. Does this sound more reasonable?

user100561
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