# Praeclarum Theorema

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## Theorem

### Formulation 1

- $\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$

### Formulation 2

- $\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)$

## Leibniz' Proof

The **praeclarum theorema** was noted and named by Leibniz, who stated and proved it in the following manner:

- If $a$ is $b$ and $d$ is $c$, then $ad$ will be $bc$.
- This is a fine theorem, which is proved in this way:
- $a$ is $b$, therefore $ad$ is $bd$ (by what precedes),
- $d$ is $c$, therefore $bd$ is $bc$ (again by what precedes),
- $ad$ is $bd$, and $bd$ is $bc$, therefore $ad$ is $bc$.
- Q.E.D.

- (Leibniz,
*Logical Papers*, p. 41).

## Linguistic Note

**Praeclarum Theorema** is Latin for **splendid theorem**.

It was so named by Gottfried Wilhelm von Leibniz.

## Also see

Compare the Constructive Dilemma, which is similar in appearance.

## Sources

- Gottfried W Leibniz:
*Addenda to the Specimen of the Universal Calculus*(1679 – 1686): pp. $40 - 46$

- John F. Sowa:
*Peirce's Rules of Inference*: Online version (2002)

- Frithjof Dau: Computer Animated Proof of Leibniz's Praeclarum Theorema (2008)

- Norman Megill: Praeclarum Theorema @ Metamath Proof Explorer (2008)