Translate English statements into propositions of predicate logic...

Question

So I tried to solve a few logic problems but I'm not sure that I translated these statements correctly...

$B(x)$: $x$ breathes fire,

$D(x)$: $x$ is a dragon,

$F(x)$: $x$ can fly,

$R(x; y)$: $x$ frightens $y$ (or $y$ is frightened by $x$),

$W(x)$: $x$ has wings (or $x$ is winged, etc.).

A. Dragon exists. $= D(x)$

B. All dragons have wings, but some cannot fly. $= \forall x [ D(x) → W(x) ] ∧ \exists x \neg F(x)$

C. Not everything is a dragon. $= \neg\forall x D(x)$

D. There is a dragon frightened by nothing. $= \neg \exists x R(x, y)$

E. For every dragon, there is a dragon that frightens it. $= \forall x [D(x) → \exists x R(x, y)]$

F. Some dragons frighten themselves. $= \exists x [ D(x) ∧ R(x, x) ]$

Answer 1

Hint: This is a very nice start. C and F are completely correct at the moment. Here are a few things in general to think about.

• All of your variables need a quantifier. When you have both $x$ and $y$ in an equation, both need to be quantified.

• Make sure that every statement that talks about dragons has a $D(\cdot)$ predicate in it.

Otherwise, try reading each of your statements and see if they precisely describe the sentence that is written for it.

Answer 2

A) No. $\exists dD(d)$

B) No. $\forall x(Dx\to Wx)\land \exists y(Dy\land\lnot Fy)$

C) Correct.

D) No. $\exists x (Dx\land \forall y\lnot Ryx)$

E) No. $\forall x(Dx\to \exists y (Dy\land Ryx))$

F) Correct.