Constructing interpretations that witness a claim: showing FOL sentences are not validities, not equivalent, or jointly satisfiable.
forall x: Calgary, Using interpretations
3 exercises
Show that each of the following is neither a validity nor a contradiction:
D(a)∧D(b)D(a)\wedge D(b)
∃x T(x,h)\exists x\,T(x,h)
P(m)∧¬∀x P(x)P(m)\wedge\neg\forall x\,P(x)
∀z J(z)↔∃y J(y)\forall z\,J(z)\leftrightarrow\exists y\,J(y)
∀x(W(x,m,n)∨∃y L(x,y))\forall x(W(x,m,n)\vee\exists y\,L(x,y))
∃x(G(x)→∀y M(y))\exists x(G(x)\rightarrow\forall y\,M(y))
∃x(x=h∧x=i)\exists x(x=h\wedge x=i)
Show that the following pairs of sentences are not logically equivalent.
J(a)J(a), K(a)K(a)
∃x J(x)\exists x\,J(x), J(m)J(m)
∀x R(x,x)\forall x\,R(x,x), ∃x R(x,x)\exists x\,R(x,x)
∃x P(x)→Q(c)\exists x\,P(x)\rightarrow Q(c), ∃x(P(x)→Q(c))\exists x(P(x)\rightarrow Q(c))
∀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))
∃x(P(x)∧Q(x))\exists x(P(x)\wedge Q(x)), ∃x(P(x)→Q(x))\exists x(P(x)\rightarrow Q(x))
∀x(P(x)→Q(x))\forall x(P(x)\rightarrow Q(x)), ∀x(P(x)∧Q(x))\forall x(P(x)\wedge Q(x))
∀x∃y R(x,y)\forall x\exists y\,R(x,y), ∃x∀y R(x,y)\exists x\forall y\,R(x,y)
∀x∃y R(x,y)\forall x\exists y\,R(x,y), ∀x∃y R(y,x)\forall x\exists y\,R(y,x)
Show that the following sentences are jointly satisfiable:
M(a),¬N(a),Pa,¬Q(a)M(a),\neg N(a),Pa,\neg Q(a)
L(e,e),L(e,g),¬L(g,e),¬L(g,g)L(e,e),L(e,g),\neg L(g,e),\neg L(g,g)
¬(M(a)∧∃x A(x)),Ma∨F(a),∀x(F(x)→A(x))\neg(M(a)\wedge\exists x\,A(x)),Ma\vee F(a),\forall x(F(x)\rightarrow A(x))
M(a)∨M(b),M(a)→∀x¬M(x)M(a)\vee M(b),M(a)\rightarrow\forall x\neg M(x)
∀y G(y),∀x(G(x)→H(x)),∃y¬I(y)\forall y\,G(y),\forall x(G(x)\rightarrow H(x)),\exists y\neg I(y)
∃x(B(x)∨A(x)),∀x¬C(x),∀x[(A(x)∧B(x))→C(x)]\exists x(B(x)\vee A(x)),\forall x\neg C(x),\forall x\bigl[(A(x)\wedge B(x))\rightarrow C(x)\bigr]
∃x X(x),∃x Y(x),∀x(X(x)↔¬Y(x))\exists x\,X(x),\exists x\,Y(x),\forall x(X(x)\leftrightarrow\neg Y(x))
∀x(P(x)∨Q(x)),∃x¬(Q(x)∧P(x))\forall x(P(x)\vee Q(x)),\exists x\neg(Q(x)\wedge P(x))
∃z(N(z)∧O(z,z)),∀x∀y(O(x,y)→O(y,x))\exists z(N(z)\wedge O(z,z)),\forall x\forall y(O(x,y)\rightarrow O(y,x))
¬∃x∀y R(x,y),∀x∃y R(x,y)\neg\exists x\forall y\,R(x,y),\forall x\exists y\,R(x,y)
¬R(a,a)\neg R(a,a), ∀x(x=a∨R(x,a))\forall x(x=a\vee R(x,a))
∀x∀y∀z[(x=y∨y=z)∨x=z]\forall x\forall y\forall z[(x=y\vee y=z)\vee x=z], ∃x∃y ¬x=y\exists x\exists y\ \neg x=y
∃x∃y((Z(x)∧Z(y))∧x=y)\exists x\exists y((Z(x)\wedge Z(y))\wedge x=y), ¬Z(d)\neg Z(d), d=ed=e
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.