Biconditional is Reflexive/Proof 1
Jump to navigation
Jump to search
Theorem
- $\vdash p \iff p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $p \implies p$ | Theorem Introduction | (None) | Law of Identity: Formulation 2 | ||
2 | $p \iff p$ | Biconditional Introduction: $\iff \II$ | 1, 1 |
$\blacksquare$