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Exercise B

Avg difficulty 3.0

Show that the following pairs of sentences are not logically equivalent.

  1. J(a)J(a), K(a)K(a)

  2. ∃x J(x)\exists x\,J(x), J(m)J(m)

  3. ∀x R(x,x)\forall x\,R(x,x), ∃x R(x,x)\exists x\,R(x,x)

  4. ∃x P(x)→Q(c)\exists x\,P(x)\rightarrow Q(c), ∃x(P(x)→Q(c))\exists x(P(x)\rightarrow Q(c))

  5. ∀x(P(x)→¬Q(x))\forall x(P(x)\rightarrow\neg Q(x)), ∃x(P(x)∧¬Q(x))\exists x(P(x)\wedge\neg Q(x))

  6. ∃x(P(x)∧Q(x))\exists x(P(x)\wedge Q(x)), ∃x(P(x)→Q(x))\exists x(P(x)\rightarrow Q(x))

  7. ∀x(P(x)→Q(x))\forall x(P(x)\rightarrow Q(x)), ∀x(P(x)∧Q(x))\forall x(P(x)\wedge Q(x))

  8. ∀x∃y R(x,y)\forall x\exists y\,R(x,y), ∃x∀y R(x,y)\exists x\forall y\,R(x,y)

  9. ∀x∃y R(x,y)\forall x\exists y\,R(x,y), ∀x∃y R(y,x)\forall x\exists y\,R(y,x)

Magnus et al., forall x: Calgary, Interpretations, Using interpretations, Exercise B

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Adapted from forall x: Calgary © 2025 P. D. Magnus, Tim Button, Robert Trueman, and Richard Zach. Licensed under CC BY 4.0. Source: forall x: Calgary. Imported at revision ec34df8.