G52PAS 2011-2012 Answer to the exercise on FOL
Express the following sentences in first-order logic:
1. A friend of a friend is a friend. (Use a binary predicate F riend.)
2. An enemy’s enemy is a friend. (Use binary predicates F riend and Enemy.)
3. If two people are friends, then they are not enemies. (Use binary predicates
F riend and Enemy.)
4. Any two people are either enemies or friends. (Use binary predicates
F riend and Enemy.)
5. John has a friend. (Use a binary predicate F riend and constant J for
John.)
6. John has at least two friends. (Use a binary predicate F riend, constant
J for John, and equality=.)
7. John has exactly two friends, and everyone else is an enemy. (Use binary
predicates F riend and Enemy, constant J for John, and equality=.)
Answer
I used F riend(x, y) for ‘y is a friend of x’ and also for ‘x and y are friends’. May
be the latter is better expressed as F riend(x, y) ∧ F riend(y, x)’, but I wanted
to keep the formulas shorter. (Same for enemies.)
1. A friend of a friend is a friend.
∀x∀y∀z(F riend(x, y) ∧ F riend(y, z) ⇒ F riend(x, z))
2. An enemy’s enemy is a friend.
∀x∀y∀z(Enemy(x, y) ∧ Enemy(y, z) ⇒ F riend(x, z))
3. If two people are friends, then they are not enemies.
∀x∀y(F riend(x, y) ⇒ ¬Enemy(x, y))
4. Any two people are either enemies or friends.
∀x∀y(F riend(x, y) ∨ Enemy(x, y))
5. John has a friend.
∃xF riend(J, x)
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6. John has at least two friends.
∃x∃y(F riend(J, x) ∧ F riend(J, y) ∧ ¬(x = y))
7. John has exactly two friends, and everyone else is an enemy.
∃x∃y(F riend(J, x)∧F riend(J, y)∧¬(x = y)∧∀z(¬(z = x)∧¬(z = y) ⇒ Enemy(J, z))
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