Theorem
- $\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$
Proof
By the tableau method of natural deduction:
$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q} $
| Line |
|
Pool
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Formula
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Rule
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Depends upon
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Notes
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| 1 |
|
1
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$\neg \paren {p \iff q}$
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Premise
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(None)
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|
| 2 |
|
1
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$\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$
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Sequent Introduction
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1
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Rule of Material Equivalence
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| 3 |
|
1
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$\neg \paren {p \implies q} \lor \neg \paren {q \implies p}$
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Sequent Introduction
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2
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De Morgan's Laws: Disjunction of Negations
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| 4 |
|
4
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$p$
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Assumption
|
(None)
|
|
| 5 |
|
4
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$p$
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Law of Identity
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4
|
|
| 6 |
|
6
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$q$
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Assumption
|
(None)
|
|
| 7 |
|
6
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$p \implies q$
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Rule of Implication: $\implies \II$
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4 – 6
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Assumption 4 has been discharged
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| 8 |
|
6
|
$\neg \neg \paren {p \implies q}$
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Double Negation Introduction: $\neg \neg \II$
|
7
|
|
| 9 |
|
1, 6
|
$\neg \paren {q \implies p}$
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Modus Tollendo Ponens $\mathrm {MTP}_1$
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3, 8
|
|
| 10 |
|
1, 6
|
$q \land \neg p$
|
Sequent Introduction
|
9
|
Conjunction with Negative is Equivalent to Negation of Conditional
|
| 11 |
|
1, 6
|
$\neg p$
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Rule of Simplification: $\land \EE_2$
|
10
|
|
| 12 |
|
1, 4, 6
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$\bot$
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Principle of Non-Contradiction: $\neg \EE$
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4, 11
|
|
| 13 |
|
1, 4
|
$\neg q$
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Proof by Contradiction: $\neg \II$
|
6 – 12
|
Assumption 6 has been discharged
|
| 14 |
|
1
|
$p \implies \neg q$
|
Rule of Implication: $\implies \II$
|
4 – 13
|
Assumption 4 has been discharged
|
| 15 |
|
15
|
$\neg q$
|
Assumption
|
(None)
|
|
| 16 |
|
15
|
$\neg q$
|
Law of Identity
|
15
|
|
| 17 |
|
17
|
$\neg p$
|
Assumption
|
(None)
|
|
| 18 |
|
17
|
$\neg q \implies \neg p$
|
Rule of Implication: $\implies \II$
|
16 – 17
|
Assumption 16 has been discharged
|
| 19 |
|
17
|
$p \implies q$
|
Sequent Introduction
|
18
|
Rule of Transposition
|
| 20 |
|
1, 17
|
$\neg \paren {q \implies p}$
|
Modus Tollendo Ponens $\mathrm {MTP}_1$
|
3, 19
|
|
| 21 |
|
1, 17
|
$q \land \neg p$
|
Sequent Introduction
|
20
|
Conjunction with Negative is Equivalent to Negation of Conditional
|
| 22 |
|
1, 17
|
$q$
|
Rule of Simplification: $\land \EE_1$
|
21
|
|
| 23 |
|
1, 15, 17
|
$\bot$
|
Principle of Non-Contradiction: $\neg \EE$
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15, 22
|
|
| 24 |
|
1, 15
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$p$
|
Proof by Contradiction: $\neg \II$
|
17 – 23
|
Assumption 17 has been discharged
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| 25 |
|
1
|
$\neg q \implies p$
|
Rule of Implication: $\implies \II$
|
15 – 24
|
Assumption 15 has been discharged
|
| 26 |
|
1
|
$\paren {p \iff \neg q}$
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Biconditional Introduction: $\iff \II$
|
14, 25
|
|
$\blacksquare$
Law of the Excluded Middle
This proof depends on the Law of the Excluded Middle.
This is one of the logical axioms that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.
However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.
This in turn invalidates this proof from an intuitionistic perspective.
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 | Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: I was tired when I posted this up so probably missed out on some obvious usage of Properties of Biconditional that would make this considerably easier. If you find such, then (unless there are blatant flaws in the above) feel free to add them as new proofs.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.
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