A survey problem asks: "In a class of 50, how many students like neither cricket nor football?" A student starts listing cases — who likes only cricket, who likes only football — and gets bogged down. Meanwhile, the problem has given them a direct gift in its wording: "neither A nor B." Three English words that translate into a specific set expression, and once translated, the arithmetic is a one-liner.
The translation is a recognition cue. The moment you read "neither ... nor ...," reach for De Morgan's first law.
The cue and its translation
- "Neither A nor B" in English = "not in A, and not in B" in logic = A' \cap B' in set notation.
- By De Morgan's first law, A' \cap B' = (A \cup B)'.
- So "neither" is the complement of the union. Two expressions, same set.
The cardinality version — the version most problems actually ask for — is:
The last equality uses inclusion-exclusion. Once you have |U|, |A|, |B|, and |A \cap B|, the count of "neither" is a single line.
Why the recognition matters
The trap is to attack "neither" directly — try to count students who are simultaneously not in A and not in B by going through the class roster. That is valid but slow. The recognition shortcut converts the "neither" statement into a union, which is what the standard formula solves.
Without the cue:
"Hmm, neither cricket nor football. So I need students who don't like cricket, but I also need them to not like football. Let me see... students who don't like cricket = 50 - 25 = 25. But some of those 25 still like football. Subtract those who like football but not cricket... which is |F| - |C \cap F| = 20 - 10 = 10. So neither cricket nor football = 25 - 10 = 15."
That path works but costs three steps of reasoning per problem.
With the cue:
"Neither cricket nor football → (C \cup F)' → |U| - |C \cup F| = 50 - (25 + 20 - 10) = 15."
Same answer, one line of arithmetic.
The three-sentence pattern for exam use
- Spot the word "neither" (or "none of" or "outside all of" for the n-set version).
- Rewrite as the complement of the union: (A \cup B)' for two sets, (A \cup B \cup C)' for three.
- Compute via |U| - |A \cup B \cup \dots| using inclusion-exclusion on the union.
Three sentences, no Venn diagram required — though drawing one on the rough sheet keeps you honest.
Walked mini-examples
Example 1. 40 students, 25 like maths (M), 18 like physics (P), 10 like both. How many like neither?
Recognise "neither" → (M \cup P)'. Count:
Answer: 7.
Example 2. 100 people surveyed. 60 use WhatsApp (W), 45 use Instagram (I), 30 use both. How many use neither?
Answer: 25.
Example 3 — three sets. 80 students, 50 play cricket (C), 40 play football (F), 30 play chess (X). 20 play C and F, 15 play F and X, 12 play C and X, 5 play all three. How many play none?
Recognise "none of the three" → (C \cup F \cup X)'. Three-set inclusion-exclusion:
Answer: 2.
Related phrasings that trigger the same translation
The word "neither" is the most common cue, but several other phrasings mean the same thing. Train your eyes to notice all of them.
| Phrasing | Set translation |
|---|---|
| "neither A nor B" | A' \cap B' |
| "not in A and not in B" | A' \cap B' |
| "outside both A and B" | A' \cap B' = (A \cup B)' |
| "in none of A, B, C" | A' \cap B' \cap C' = (A \cup B \cup C)' |
| "students who did not pass any subject" | complement of the union over all subjects |
| "customers who bought nothing from our three brands" | complement of the union over all brands |
Every row in this table collapses to a complement-of-union, which is then counted via |U| - |\text{union}| using inclusion-exclusion.
The mirrored cue: "both" triggers intersection, not union
Worth flagging the neighbour pattern so you do not confuse them.
- "Neither A nor B" \to A' \cap B' = (A \cup B)' (complement of union).
- "Both A and B" \to A \cap B (intersection).
- "At least one of A, B" \to A \cup B (union).
- "Exactly one of A, B" \to A \triangle B (symmetric difference).
Four phrasings, four different set expressions. The "neither → (union)'" translation is the one students forget most often because it requires two steps (first to A' \cap B', then to (A \cup B)' via De Morgan) — but once the recognition is reflex, the three-sentence pattern handles every such problem.
The exam reflex
Whenever a word-problem uses "neither," "none of," "outside all of," or "not in any" — do not try to count the complement region directly. Instead:
- Write the set expression as a complement of a union.
- Apply |U| - |\text{union}|.
- Compute the union via inclusion-exclusion from the given intersection data.
This reflex converts a vague-feeling English sentence into a specific arithmetic target, and the target is always one line away. The De Morgan step is what makes the conversion legitimate — you are not guessing, you are invoking an identity that guarantees (A \cup B)' = A' \cap B'. The identity is the bridge, and "neither" is the signpost pointing to it.
Related: Set Operations · Spot the Universal Set U · Inclusion-Exclusion vs Direct Venn Counting · A − B vs A ∩ B′