MHB Understanding the Logic of Quantified Statements

[stan1992]

∃x∀y∀z[(F(x, y)∧G(x,z)) → H(y,z)]

 

[Ackbach]

\begin{align*}
\neg\big[(\exists \, x)(\forall \, y)(\forall \, z)[(F(x,y) \land G(x,z)) \to H(y,z)]\big]
&\equiv (\forall \, x)\neg\big[(\forall \, y)(\forall \, z)[(F(x,y) \land G(x,z)) \to H(y,z)]\big] \\
&\equiv (\forall \, x)(\exists \, y)\neg\big[(\forall \, z)[(F(x,y) \land G(x,z)) \to H(y,z)]\big] \\
&\equiv (\forall \, x)(\exists \, y)(\exists \, z)\neg\big[(F(x,y) \land G(x,z)) \to H(y,z)\big].
\end{align*}
Can you finish?

 

[stan1992]

≡(∀x)(∃y)(∃z)[¬(F(x,y)∧G(x,z))→¬H(y,z)]
≡(∀x)(∃y)(∃z)[¬F(x,y)∨¬G(x,z)→¬H(y,z)]

Is this correct?

 

[Evgeny.Makarov]

The negation of $A\to B$ is $A\land\neg B$ and not $\neg A\to\neg B$. In fact, it is not entirely correct to say "the negation" because each formula has infinitely many formulas equivalent to it. For the same reason, the original problem is not well-posed. As it is stated now, it is enough to add $\neg$ to the beginning the formula.

 

[stan1992]

The negation of $A\to B$ is $A\land\neg B$ and not $\neg A\to\neg B$. In fact, it is not entirely correct to say "the negation" because each formula has infinitely many formulas equivalent to it. For the same reason, the original problem is not well-posed. As it is stated now, it is enough to add $\neg$ to the beginning the formula.

≡(∀x)(∃y)(∃z)[¬F(x,y)VG(x,z))∧H(y,z)]

Would this be it?

 

[Evgeny.Makarov]

≡(∀x)(∃y)(∃z)[¬F(x,y)VG(x,z))∧H(y,z)]

Would this be it?

No. For one, this formula has unbalanced parentheses.

You should apply the law about the negation of an implication that I wrote more carefully.

 

[stan1992]

No. For one, this formula has unbalanced parentheses.

You should apply the law about the negation of an implication that I wrote more carefully.

(∀x)(∃y)(∃z)(¬F(x,y)V¬G(x,z)∧¬H(y,z))

¬G is because of Dem. and ¬H from the Def→ right?

 

[Evgeny.Makarov]

The problem is in finding an equivalent formula for $\neg((F(x, y)\land G(x,z))\to H(y,z))$. I stated that $\neg(A\to B)\equiv A\land\neg B$. Please compare two formulas:
\begin{align}
&\neg((F(x, y)\land G(x,z))\to H(y,z))\\
&\neg(A\to B)
\end{align}
What should be substituted for $A$ and $B$ so that these formulas become equal, character-by-character? After you determine this, please write what $A\land\neg B$ looks like for those concrete $A$ and $B$.

 

[stan1992]

The problem is in finding an equivalent formula for $\neg((F(x, y)\land G(x,z))\to H(y,z))$. I stated that $\neg(A\to B)\equiv A\land\neg B$. Please compare two formulas:
\begin{align}
&\neg((F(x, y)\land G(x,z))\to H(y,z))\\
&\neg(A\to B)
\end{align}
What should be substituted for $A$ and $B$ so that these formulas become equal, character-by-character? After you determine this, please write what $A\land\neg B$ looks like for those concrete $A$ and $B$.

≡(∀x)(∃y)(∃z)[(F(x,y)∧G(x,z))∧¬H(y,z)]

 

[Evgeny.Makarov]

≡(∀x)(∃y)(∃z)[(F(x,y)∧G(x,z))∧¬H(y,z)]

This is correct.