Definition:Constructed Semantics/Instance 3/Factor Principle

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Theorem

The Factor Principle:

$\left({p \implies q}\right) \implies \left({\left({r \lor p}\right) \implies \left ({r \lor q}\right)}\right)$

is a tautology in Instance 3 of constructed semantics.


Proof

By the definitional abbreviation for the conditional:

$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$

the Factor Principle can be written as:

$\neg \left({\neg p \lor q}\right) \lor \left({\neg \left({r \lor p}\right) \lor \left ({r \lor q}\right)}\right)$

This evaluates as follows:

$\begin{array}{|ccccc|c|cccccccc|} \hline \neg & (\neg & p & \lor & q) & \lor & (\neg & (r & \lor & p) & \lor & (r & \lor & q)) \\ \hline 0 & 2 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 0 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 0 & 0 \\ 0 & 2 & 2 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 0 & 1 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 2 & 0 & 2 & 2 & 1 & 0 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 1 \\ 0 & 2 & 0 & 0 & 1 & 0 & 2 & 1 & 0 & 0 & 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 2 & 0 & 2 & 2 & 1 & 0 & 0 & 1 & 2 & 2 & 0 & 1 & 1 & 1 \\ 0 & 2 & 0 & 0 & 1 & 0 & 2 & 2 & 0 & 0 & 2 & 2 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 2 & 1 \\ 2 & 0 & 2 & 2 & 1 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 2 & 1 \\ 0 & 2 & 0 & 2 & 2 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 1 & 1 & 2 & 2 & 0 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 2 \\ 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & 2 & 2 & 0 & 2 & 1 & 0 & 0 & 2 & 1 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 0 & 1 & 1 & 1 & 1 & 2 & 1 & 2 & 2 \\ 2 & 0 & 2 & 0 & 2 & 0 & 0 & 1 & 2 & 2 & 0 & 1 & 2 & 2 \\ 0 & 2 & 0 & 2 & 2 & 0 & 2 & 2 & 0 & 0 & 2 & 2 & 2 & 2 \\ 0 & 1 & 1 & 2 & 2 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 2 & 2 \\ 2 & 0 & 2 & 0 & 2 & 0 & 0 & 2 & 2 & 2 & 0 & 2 & 2 & 2 \\ \hline \end{array}$

$\blacksquare$