stan1992
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∃x∀y∀z[(F(x, y)∧G(x,z)) → H(y,z)]
Evgeny.Makarov said: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.
No. For one, this formula has unbalanced parentheses.stan1992 said:≡(∀x)(∃y)(∃z)[¬F(x,y)VG(x,z))∧H(y,z)]
Would this be it?
Evgeny.Makarov said:No. For one, this formula has unbalanced parentheses.
You should apply the law about the negation of an implication that I wrote more carefully.
Evgeny.Makarov said: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 $B$.
This is correct.stan1992 said:≡(∀x)(∃y)(∃z)[(F(x,y)∧G(x,z))∧¬H(y,z)]