Peirce's Law is Equivalent to Law of Excluded Middle
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Theorem
- $\paren {p \implies q} \implies p \vdash p$
is logically equivalent to the Law of Excluded Middle:
- $\vdash p \lor \neg p$
That is, Peirce's Law holds if and only if the Law of Excluded Middle holds.
Proof
Law of Excluded Middle implies Peirce's Law
Let the truth of the Law of Excluded Middle be assumed.
Then:
- $\paren {p \lor \neg p} \vdash \paren {\paren {p \implies q} \implies p} \implies p$
is demonstrated, as follows.
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \lor \neg p$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 2 | $\paren {\paren {p \implies q} \implies p} \implies p$ | Sequent Introduction | 2 | True Statement is implied by Every Statement | |
4 | 4 | $\neg p$ | Assumption | (None) | ||
5 | 4 | $p \implies q$ | Sequent Introduction | 4 | False Statement implies Every Statement | |
6 | 6 | $\paren {p \implies q} \implies p$ | Assumption | (None) | ||
7 | 6, 5 | $p$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 6, 5 | ||
8 | 5 | $\paren {\paren {p \implies q} \implies p} \implies p$ | Rule of Implication: $\implies \II$ | 6 – 7 | Assumption 6 has been discharged | |
9 | 1 | $\paren {\paren {p \implies q} \implies p} \implies p$ | Proof by Cases: $\text{PBC}$ | 1, 2 – 3, 4 – 8 | Assumptions 2 and 4 have been discharged |
The result follows by an application of Modus Ponendo Ponens:
- $\paren {\paren {p \implies q} \implies p} \implies p, \paren {p \implies q} \implies p \vdash p$
$\Box$
Peirce's Law implies Law of Excluded Middle
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $(p \lor \neg p) \implies \bot$ | Assumption | (None) | ||
2 | 1 | $\neg(p \lor \neg p)$ | Sequent Introduction | 1 | Negation as Implication of Bottom | |
3 | 1 | $\bot$ | Sequent Introduction | 2 | Negation of Excluded Middle is False | |
4 | 1 | $p \lor \neg p$ | Rule of Explosion: $\bot \EE$ | 3 | ||
5 | $((p \lor \neg p) \implies \bot) \implies p \lor \neg p$ | Rule of Implication: $\implies \II$ | 1 – 4 | Assumption 1 has been discharged | ||
6 | $p \lor \neg p$ | Sequent Introduction | 5 | Peirce's Law |
$\blacksquare$