Factor Principles

Theorem

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


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

They can alternatively be rendered as:


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


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

The forms can be seen to be logically equivalent by application of the Rule of Implication and Modus Ponendo Ponens.

Proof

 * align="right" | 2 ||
 * align="right" |
 * $r \implies r$
 * Law of Identity
 * (None)
 * This is a theorem so depends on nothing.
 * This is a theorem so depends on nothing.


 * align="right" | 4 ||
 * align="right" | 1
 * $\left({p \land r}\right) \implies \left ({q \land r}\right)$
 * Sequent Introduction
 * 3
 * Praeclarum Theorema
 * Praeclarum Theorema

And the second is like it, namely this:


 * align="right" | 2 ||
 * align="right" |
 * $r \implies r$
 * Law of Identity
 * (None)
 * This is a theorem so depends on nothing.
 * This is a theorem so depends on nothing.


 * align="right" | 4 ||
 * align="right" | 1
 * $\left({r \land p}\right) \implies \left ({r \land q}\right)$
 * Sequent Introduction
 * 3
 * Praeclarum Theorema
 * Praeclarum Theorema

It would, of course, be possible to derive the second from the first by applying the Rule of Commutation to the conjunctions on the RHS, but this is unnecessarily fiddly for a result so obvious. The Praeclarum Theorema does all the work we need instead.