Praeclarum Theorema/Formulation 1

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \land \left({r \implies s}\right)$ Premise (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $r \implies s$ Rule of Simplification: $\land \mathcal E_2$ 1
4 4 $p \land r$ Assumption (None)
5 4 $p$ Rule of Simplification: $\land \mathcal E_1$ 4
6 4 $r$ Rule of Simplification: $\land \mathcal E_2$ 4
7 1, 4 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 5
8 1, 4 $s$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 6
9 1, 4 $q \land s$ Rule of Conjunction: $\land \mathcal I$ 7, 8
10 1 $\left({p \land r}\right) \implies \left({q \land s}\right)$ Rule of Implication: $\implies \mathcal I$ 4 – 9 Assumption 4 has been discharged

$\blacksquare$


Proof 2

We apply the Method of Truth Tables to the proposition.

As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the left hand side is $T$, that under the one on the right hand side is also $T$:


$\begin{array}{|ccccccc||ccccccc|} \hline (p & \implies & q) & \land & (r & \implies & s) & (p & \land & r) & \implies & (q & \land & s) \\ \hline F & T & F & T & F & T & F & F & F & F & T & F & F & F \\ F & T & F & T & F & T & T & F & F & F & T & F & F & T \\ F & T & F & F & T & F & F & F & F & T & T & F & F & F \\ F & T & F & T & T & T & T & F & F & T & T & F & F & T \\ F & T & T & T & F & T & F & F & F & F & T & T & F & F \\ F & T & T & T & F & T & T & F & F & F & T & T & T & T \\ F & T & T & F & T & F & F & F & F & T & T & T & F & F \\ F & T & T & T & T & T & T & F & F & T & T & T & T & T \\ T & F & F & F & F & T & F & T & F & F & T & F & F & F \\ T & F & F & F & F & T & T & T & F & F & T & F & F & T \\ T & F & F & F & T & F & F & T & T & T & F & F & F & F \\ T & F & F & F & T & T & T & T & T & T & F & F & F & T \\ T & T & T & T & F & T & F & T & F & F & T & T & F & F \\ T & T & T & T & F & T & T & T & F & F & T & T & T & T \\ T & T & T & F & T & F & F & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

Hence the result.


Note that the two formulas are not equivalent, as the relevant columns do not match exactly.

$\blacksquare$


Proof 3

By the tableau method of natural deduction:

$\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \land \left({r \implies s}\right)$ Premise (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $r \implies s$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $p \land r \implies q \land r$ Sequent Introduction 2 Factor Principles/Conjunction on Right/Formulation 1/Proof 2
5 1 $q \land r \implies q \land s$ Sequent Introduction 3 Factor Principles/Conjunction on Left/Formulation 1/Proof 2
6 1 $p \land r \implies q \land s$ Sequent Introduction 4,5 Hypothetical Syllogism

$\blacksquare$


Sources