Symbolic LogicRules of Replacement

De Morgan and Double Negation (DM, DN)

DM: ¬(p ∧ q) :: ¬p ∨ ¬q and ¬(p ∨ q) :: ¬p ∧ ¬q. DN: p :: ¬¬p.

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The first two rules of replacement are De Morgan's rule (DM) and double negation (DN). You know both equivalences already from De Morgan's Laws; now they become steps in a proof. They are the rules from Rules of Replacement you will use most.

Each row: the two forms may replace each other anywhere.
RuleFormForm
DM$\neg (p \land q)$$\neg p \lor \neg q$
DM$\neg (p \lor q)$$\neg p \land \neg q$
DN$p$$\neg \neg p$

Key idea

DM moves a negation into or out of a conjunction or disjunction, switching the connective. DN adds or removes two negations in front of the same formula. Both may be used on any part of a line, in either direction, one application per line.

Why it works

Both pairs are equivalent by truth table, so either form can stand in for the other. DN is the quiet helper: rules like MT and DS need an exact negation, and DN supplies it. From $p \to \neg q$ and $q$, MT needs $\neg \neg q$, the negation of the consequent $\neg q$, and DN turns $q$ into $\neg \neg q$.
DN fixes the form so MT applies.

Worked examples

Worked example: DM on a whole line

Apply DM to $\neg (r \lor s)$.
  1. A negated disjunction becomes a conjunction of negations.

Answer: $\neg r \land \neg s$

Worked example: DM inside a line

Apply DM to the antecedent of $\neg (p \land q) \to r$.
  1. Only the antecedent changes.

Answer: $(\neg p \lor \neg q) \to r$

Worked example: DN removing negations

Apply DN to $\neg \neg t \land u$.
  1. Remove the two negations in front of $t$.

Answer: $t \land u$

Worked example: Both rules in a proof

Premises $\neg (p \lor q)$ and $r \to p$. Prove $\neg r$.
  1. Line 3: $\neg p \land \neg q$ by DM 1.
  2. Line 4: $\neg p$ by Simp 3.
  3. Line 5: $\neg r$ by MT 2, 4.

Answer: Three lines.

Common mistake

Moving the negation inside but keeping the connective: writing $\neg (p \lor q)$ as $\neg p \lor \neg q$. That is not DM; it changes the meaning (check the row $p$ T, $q$ F). DM always switches $\land$ and $\lor$. A second slip: using DN on two negations that cover different parts, such as turning $\neg (\neg p \land q)$ into $p \land q$. The two negations must stand directly in front of the same formula.

When should you reach for DM? Whenever a negation stands in front of parentheses and the next rule needs the parts separately. A negated conjunction becomes a disjunction, which DS can use; a negated disjunction becomes a conjunction, which Simp can take apart. DN is the finishing touch whenever a rule wants $\neg \neg p$ and you have $p$, or the other way round.

Typing tip: in the answer box, ~~p is read as $\neg \neg p$. There is no need for parentheses around a double negation.

Each use of a replacement rule changes one part, once. If a line needs DM in two places, use two lines, each justified "DM" with the line it came from. That keeps every step easy to check.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Builds on

Leads to

Sources and further reading

  • Patrick J. Hurley and Lori Watson, A Concise Introduction to Logic, 13th ed., section 7.3
  • forall x: Calgary, an open introduction to formal logic (P. D. Magnus, Tim Button and others; Open Logic Project), chapter on derived rules and replacement
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