PRELIMINARY

9: Prove the symmetry of inequality, $\forall{x}\forall{y}(x=y\rightarrow y=x)$

I'm pretty confused about the way to use the formal logic axioms and structural rules, but this is my best shot.

1. $\vdash x=x, \vdash y=y$ by the axiom of equality.

2. $\vdash x=x \wedge y=y$ by and introduction from 1.

3. $(x=x \wedge y=y), x=y \vdash y=x$…. by???? substitution (is that something we can use?) from 2.

4. $(x=x \wedge y=y) \vdash x=y \rightarrow y=x$ by $\rightarrow$ introduction.

5. $\vdash x=y \rightarrow y=x$ by (however you say that $\Phi \vdash \Sigma, \Sigma \vdash \Psi$ proves that $\Phi \vdash \Psi)$ and 2 and 4.

6. $\vdash \forall{y}(x=y \rightarrow y=x)$ by $\forall$ introduction from 5.

7. $\vdash \forall{x}\forall{y}(x=y \rightarrow y=x)$ by $\forall$ introduction from 6.

10a: (Again, really rough here): Prove that $\exists{x}(x=x)$.

1. $\vdash (x=x)$ by the axiom of equality.

2. $\vdash \forall{x}(x=x)$ by $\forall$ introduction and 1.

3. $\vdash (x=x)[y|x]$ by $\forall$ elimination and 2.

4. $\vdash \exists{x}(x=x)$ by $\exists$ introduction and 3.

10b: Write an unofficial wff for there are two things:

$\forall{x}\exists{y}\lnot (x = y)$

FINALS

9. Prove the symmetry of inequality, $\forall{x}\forall{y}(x=y\rightarrow y=x)$

Really no clue how to get here… something using like $\{x=y, \Phi{(y)}\}\vdash \Phi{(x)}$, then $x=y \vdash \Phi{(y)} \rightarrow \Phi{(x)}$. Anyway, once we get to the following step….

$x=y \vdash y=x$

$\vdash(x=y\rightarrow y=x)$ by $\rightarrow$I and the previous.

$\vdash\forall{y}(x=y\rightarrow y=x)$ by $\forall$I and the previous.

$\vdash\forall{x}\forall{y}(x=y\rightarrow y=x)$ by $\forall$I and the previous.

10a. Prove that $\vdash\exists{x}(x=x)$.

1. $\vdash (x=x)$ by the axiom of equality.

2. $\vdash \forall{x}(x=x)$ by $\forall$ introduction and 1.

3. $\vdash (x=x)[y|x]$ by $\forall$ elimination and 2.

4. $\vdash \exists{x}(x=x)$ by $\exists$ introduction and 3.

10b: Write an unofficial wff for there are two things:

$\forall{x}\exists{y}\lnot (x = y)$

Comments - Andrew Furash

You have the end of the first proof, but you need to use the =E rule to get the first part. Great job on 10. I think 10B is correct but you could always have written:

$\exists{x}\exists{y}\lnot (x=y)$