there’s going to be a certain amount of fiddling around right in here. of course there’s a *better* way but what we’re after is the *easy* way. standards are great that’s why there’s so many *of* ’em.

\equiv\leftrightarrow\Leftrightarrow.

any luck? three different symbols, right?
an equal-sign-that-didn’t-know-when-to-stop;
a “left-and-right-arrow”; and a *double* leftrightarrow.
right?

okay. it appears to work on *this* set-up;
no telling what *your* mileage looks like
(just be sure that it *will* vary).

these symbols, then.

these symbols *all* stand for versions of
“two things have the same interpretation”.

there’s already a great deal of confusion
about (what i take to be) the *best-known*
symbol in this class: the *sign of equality*, “=”.
i’ve remarked on this phenomenon before.

but at the “math 366” level–my current class–
it becomes very important to be very careful to
*distinguish* between various notions of
“means the same thing as”.

but… how?

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  1. \centerline{Math 366}
    \centerline{Quiz 2}
    \vskip .5in
    \parindent=0

    {\bf 1.}
    Each of the arguments
    {\it (a)} through {\it (d)}
    is VALID or NOT VALID;
    determine which.
    For VALID arguments,
    name the valid argument
    {\it form} (Modus Ponens or
    Transitivity,
    for example, or [be careful!]
    their Universal counterparts).
    For arguments that are
    NOT VALID, name the
    fallacy illustrated
    (Converse Error
    or Inverse Error,
    for example).

    \vfil
    {\it (a)}

    If that’s a thermometer, then I’m Bugs Bunny.

    That’s {\it not} a thermometer.

    $\bullet$ I’m not Bugs Bunny.

    \vfil
    {\it (b)}

    $t$ is rational or $t$ is irrational.

    $t$ is not irrational.

    $\bullet t$ is rational.

    \vfil
    {\it (c)}

    $(\forall x \in R) (x \not\in Q}) \rightarrow (x \in \^Q)$

    $\sim(t \in \^Q)$

    $\bullet \sim(t \not\in Q)$

    \vfil
    {\it (d)}

    If Pigs Fly, then there’s a Sty in the Sky.

    There {\it is} a Sty in the Sky.

    $\bullet$ Pigs Fly.
    \vfil\vfil
    {\bf 2.}
    Use BOTH of {\it (a)} a Truth Table AND {\it (b)} a sequence
    of Equivalences (from the textbook-and-handout list… naming
    a justification for each “step”) to prove
    the equivalence $p\wedge \sim(q \wedge \sim p) \equiv p$.

    \vfil\vfil\vfil\vfil
    \vfil\vfil\vfil\vfil\vfil\vfil
    \vfil\vfil\vfil\vfil
    \vfil\vfil\vfil\vfil\vfil\vfil
    \vfil\vfil
    \vfil\eject

    {\bf 3.}Is Owen in debt?

    Create symbols appropriately
    (“OM” replacing “Owen is a millionaire”,
    for example). Think through the problem
    to discover its Answer.
    Then {\it justify} your answer with
    a sequence of “steps”
    written in these symbols
    (and appropriate “logic” code);
    each consisting of a three-line
    (major premise, minor premise,
    conclusion) “argument”. Each
    {\it premise} of a given step
    should come from the given
    information ({\it (a)}–{\it (e)})
    or a previous “step”; each
    {\it conclusion} will (of course!)
    be an instance of some valid
    Rule of Inference.
    (Compare “Example 2.3.8″ at
    Epp, {\it pp.} 56{\it f}).

    {\it a.} Owen is a millionaire or Owen is underinsured.

    {\it b.} If Owen is underinsured, then Owen is in debt.

    {\it c.} If Owen is a millionaire, then Owen is {\it not} in debt.

    {\it d.} If Owen is a millionaire, Owen has quit his job.

    {\it e.} Owen has not quit his job.
    \vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil\vfil

    {\bf 4.}

    Write appropriate negations.
    It does {\it not} suffice simply to
    affix “$\sim$” (or the phrase
    “it is not the case that…”
    [for example]); the negation
    of $x<17$ isn't $\sim(x<17)$,
    for example [or $x\not<17$
    for that matter], but $x \ge 17$.
    \vfil
    $\bullet (\forall x) P(x) \rightarrow Q(x)$
    \vfil
    $\bullet x<0$ or $x\ge 1$
    \vfil
    $\bullet$ There ain't no life nowhere.

    \vfil\eject

    \bye

  2. http://pdfcast.org/pdf/366q2-pdf




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