Praeclarum Theorema/Formulation 1/Proof 3
Jump to navigation
Jump to search
Theorem
- $\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \implies q} \land \paren {r \implies s}$ | Premise | (None) | ||
2 | 1 | $p \implies q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $r \implies s$ | Rule of Simplification: $\land \EE_2$ | 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 |
$\blacksquare$