Equality is Transitive
From ProofWiki
Theorem
Equality is transitive.
That is:
- $\forall a, b, c: \left({a = b}\right) \land \left({b = c}\right) \implies a = c$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle a\) | \(=\) | \(\displaystyle b\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \vdash\) | \(\displaystyle \) | \(\displaystyle P \ \left({a}\right)\) | \(\iff\) | \(\displaystyle P \ \left({b}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Leibniz's Law | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle b\) | \(=\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \vdash\) | \(\displaystyle \) | \(\displaystyle P \ \left({b}\right)\) | \(\iff\) | \(\displaystyle P \ \left({c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Leibniz's Law | ||
| \(\displaystyle \) | \(\displaystyle \vdash\) | \(\displaystyle \) | \(\displaystyle P \ \left({a}\right)\) | \(\iff\) | \(\displaystyle P \ \left({c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Biconditional is Transitive | ||
| \(\displaystyle \) | \(\displaystyle \vdash\) | \(\displaystyle \) | \(\displaystyle a\) | \(=\) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Leibniz's Law |
$\blacksquare$
Sources
- Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 1$: The Axiom of Extension