Formal Logic Flashcards

1
Q

Everything caused something.

A

∀x∃yCxy

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2
Q

Everything is caused by something.

A

∀y∃xCxy

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3
Q

Something caused everything.

A

∃x∀yCxy

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4
Q

Something is caused by everything.

A

∃y∀xCxy

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5
Q

Everything caused everything.

A

∀x∀yCxy

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6
Q

Something caused something.

A

∃x∃yCxy

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7
Q

Everything caused itself.

A

∀xCxx

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8
Q

Something caused itself.

A

∃xCxx

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9
Q
  1. Mary is taller than John.
  2. John is taller than Lucy.
  3. Whoever is taller than someone who is taller than a third person, is taller than this third person.
    C. Mary is taller than Lucy.
A
  1. Tmj
  2. Tjl
  3. ∀x∀y∀z((Txy ∧ Tyz) → Txz)
  4. Tml
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10
Q

What is logically equivalent to ∃xFx ?

A

¬∀x¬Fx

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11
Q

What is logically equivalent to ∀xFx

A

¬∃x¬Fx

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12
Q

What does ¬∀x¬Fx mean?

A

‘It is not the case that everything is not-F’

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13
Q

What does ¬∃x¬Fx mean?

A

Nothing is not-F

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14
Q

Polar bears and grizzly bears are dangerous

A

Anything which is either a polar bear or a grizzly bear is
dangerous:
∀x((Gx ∨ Px) → Dx)

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15
Q

There is at least one unicorn

A

∃xUx

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16
Q

There is at most one unicorn

A

If you come across a unicorn and then you come across a unicorn again, they are identical.
If x is a unicorn and y is a unicorn, they must be identical.
∀x∀y((Ux ∧ Uy) → x = y)

17
Q

There is exactly one unicorn.

A

If you encounter this one unicorn, and you come across another one, that latter unicorn must be identical to the first.
There is a unicorn and anything that is a unicorn is identical to it.
∃x(Ux ∧ ∀y(Uy → x = y))

18
Q

There are at least two unicorns.

A

There are x and y that are unicorns and they are different.

∃x∃y((Ux ∧ Uy) ∧ x ̸= y)

19
Q

There are at most two unicorns.

A

If x, y and z are unicorns, then one of them must be identical to another one of them.
∀x∀y∀z(((Ux ∧ Uy) ∧ Uz) → (x = y ∨ x = z))

20
Q

There are exactly two unicorns.

A

There are different x and y that are unicorns, for any z that is a unicorn, it is identical to one of them.
∃x∃y(((Ux ∧ Uy) ∧ x ̸= y) ∧ ∀z(Uz → (z = x ∨ z = y)))

21
Q

The present King of France is bald.

A

There is exactly one present King of France and he is bald.

∃x((Fx ∧ ∀y(Fy → x = y)) ∧ Bx)

22
Q

The present King of France is not bald.

A

It is not the case that there is exactly one present King of France and he is bald.
¬∃x((Fx ∧ ∀y(Fy → x = y)) ∧ Bx)

There is exactly one present King of France and he is not bald.
∃x((Fx ∧ ∀y(Fy → x = y)) ∧ ¬Bx)

23
Q

Universal Instantiation

A

∀xF x → F m

24
Q

Existential Generalisation

A

Fm → ∃xF x