### Quiz of October 14, 2010

1. Classify the arguments.

Specifically, each is VALID or NOT VALID; determine which (explicitly; “V” or “NV” is OK).

For VALID arguments, name the valid argument form (from the given list or (be careful!) certain “Universal” generalizations of arguments from the list).

For arguments that are NOT VALID, name the fallacy illustrated (Converse Error or Inverse Error, for example).

(a)
If you don’t see sharp, you will be flat.
You will be flat.
$\bullet$ You don’t see sharp.

(b)
All mathematicians are reasonable.
My cat is not reasonable.
$\bullet$ My cat is not a mathematician.

(c)
I will die or I will pay taxes.
If I pay taxes then Big Brother is watching.
If I die then Big Brother is watching.
$\bullet$ Big Brother is watching.

(d)
If you let me get a word in edgewise, I will go on.
You will not let me get a word in edgewise.
$\bullet$ I will not go on.

(e)
I’m Jack and I’m loud.
$\bullet$ I’m loud.

(f)
If I didn’t know any better, I’d swear that’s a UFO.
I didn’t know any better.
$\bullet$ I’d swear that’s a UFO.

2.

(a.) Write out a Truth Table (a “complete” Truth Table… one column for each variable and one for each calculation) for the statement form $p \wedge \sim(q \wedge \sim p)\,.$ (You will need six columns.)

(b.) Mark the appropriate parts of the table to complete the “truth-table proof” of the equivalence $p \wedge \sim(q \wedge \sim p) \equiv p\,.$

(c.) Write out an English sentence explaining why this diagram proves the equivalence valid.

3. Use a sequence of “steps” from the given list of Logical Equivalences to show that $p \wedge \sim(q \wedge \sim p) \equiv p\,.$ Explicitly name the equivalence at each step.

4. Will all be well?

Create symbols appropriately (“ABW” replacing “All will be well”, for example). Think through the problem to discover its Answer. Then justify your answer with a sequence of “steps”; give an explicit reason for each of your steps (premises, previous steps, or “rules of inference” from the given list).

a. I will keep turning the crank.

b. I’ll get paid soon or I’ll know-the-reason-why.

c. If I’ll know-the-reason-why then all will not be well.

d. If I get paid soon then all will be well.

e. If I keep turning the crank then I’ll get paid soon.

5. Write appropriate negations, simplifying appropriately. (Recall, or example, that the negation of $x > 0$ is $x\le 0$ and that the negation of an “implication” is a “conjunction”).

$\bullet (\exists x) P(x) \wedge Q(x)$

$\bullet\, n<-5$ or $n\ge 17$

$\bullet$ If you know what's good for you then you keep your mouth shut.

$\bullet (\forall x \exists y): y+1 = x$

6. (a.) Write down a Boolean Expression for the given circuit. (OOPS NO DRAWING.)

(b.) Write down a Boolean Expression for the given Input-Output table.

T&T&T&T\cr
T&T&F&T\cr
T&F&T&F\cr
T&F&F&T\cr
F&T&T&T\cr
F&T&F&T\cr
F&F&T&F\cr
F&F&F&T\cr

LOOKS WAY PRETTIER WHEN TeX’ED

1. (a)
If you don’t see sharp, you will be flat.
You will be flat.
You don’t see sharp.

CONVERSE ERROR. (“NV”… Not Valid)
One may diagram the situation as
$\sim SS \rightarrow BF$
$BF$
$\bullet \sim SS$;
on this model one has “moved in the wrong direction” along the “arrow”.

(b)
All mathematicians are reasonable.
My cat is not reasonable.
My cat is not a mathematician.

VALID BY MODUS TOLENS
MT is the name of the “form”
$M \rightarrow R$
$\sim R$
$\bullet \sim M$
thus “MT” depends in some sense on
the equivalence of an implication with its contrapositive:
$(M \rightarrow R) \equiv (\sim R \rightarrow \sim M)$.
(… now just “apply modus ponens”.)

(c)
I will die or I will pay taxes.
If I pay taxes then Big Brother is watching.
If I die then Big Brother is watching.
Big Brother is watching.

VALID ALAS BY DIVISION-INTO-CASES
No need to formalize, right?

(d)
If you let me get a word in edgewise, I will go on.
You will not let me get a word in edgewise.
I will not go on.

INVERSE ERROR (NV)

(e)
I’m Jack and I’m loud.
I’m loud.

DAMNED IF I REMEMBER; WHERE’S THAT LIST?
Particularization from a Conjunction, let’s call it. Yeah, that’s it. My story; sticking to it (for now). Long live “pee and cue implies pee”!

(f)
If I didn’t know any better, I’d swear that’s a UFO.
I didn’t know any better.
I’d swear that’s a UFO.

VALID BY GOOD OLD FASHIONED MODUS PONENS
DNB \implies SUFO
DNB
*SUFO
One has “followed the arrow” in the right direction… this is what implications (and their “arrows”) are for!

2. 3. Use a sequence of “steps” from the given list of Logical Equivalences to show that $p \wedge \sim(q \wedge \sim p) \equiv p\,.$ Explicitly name the equivalence at each step.

No matter how you beg ’em… or punish ’em with points off… some students will persist in copying out the whole equation for their first line and working downward from there. Cf. every Trig class ever.

$p \wedge \sim(q \wedge \sim p)$
$\equiv p \wedge (\sim q \vee \sim \sim p)$
$\equiv p \wedge (\sim q \vee p)$
$\equiv p \wedge (p \vee \sim q)$
$\equiv p$

… by, respectively, De Morgan’s Laws, Double Negation, Commutativity, and an Absorbing Law.

The other main points-losing-tendency is to “do two steps at once”… for example, a student might jump from the antepenultimate line to the end (overlooking the need… with “rules” on the chart of the day… to “commute” the p and q before “applying the absorbing law”). I suppose the main point to be made here is that “the right answer” does depend on “the given list of… Eq’nces”. Sorry I haven’t bothered to include the actual list.

4. Will all be well?
a. I will keep turning the crank.
b. I’ll get paid soon or I’ll know-the-reason-why.
c. If I’ll know-the-reason-why then all will not be well.
d. If I get paid soon then all will be well.
e. If I keep turning the crank then I’ll get paid soon.

Let ABW, KTC, GPS, and KRW denote “All will be well”, “I will keep turning the crank”, “I’ll get paid soon”, and “I’ll know-the-reason-why” (respectively).

(i.)
KTC (a.)
KTC \implies GPS (e.)
* GPS (Modus Ponens)

(ii.)
GPS (i)
GPS \implies ABW (d.)
* ABW (Modus Ponens)

All will be well.

5. Write appropriate negations, simplifying appropriately. (Recall, or example, that the negation of $x > 0$ is $x\le 0$ and that the negation of an “implication” is a “conjunction”).

$\bullet (\exists x) P(x) \wedge Q(x)$
$\sim[(\exists x) P(x) \wedge Q(x)]$
$\equiv (\forall x) \sim P(x) \vee \sim Q(x)$

The negation of an existential statement… “there exists x”, $\exists x$… is a universal statement: “for all x”, $\forall x$. A De Morgan law also plays a part.

$\bullet\, n<-5$ or $n\ge 17$

The negation here is “$n\ge -5$ and $n < 17$“.

But actually I’d probably prefer $n \in [-5, 17)$ if I had my way. Trouble is, it obscures the DeMorgan (“the negation of an “or” is the “and” of the negations” or what have you).

$\bullet$ If you know what's good for you then you keep your mouth shut.

You know what’s good for you and you don’t keep your mouth shut.

Students fly right by that remark up there to the effect that “the negation of an implication is a conjunction” as if I’d said “teachertalk teachertalk mumblemumble mo”; it’d drive me nuts if somehow the calluses wore out all at once. It’s incredible. Really. “Uh-oh… technical terms!… better skip that part…” when, for hecksake, these are our very subject matter!

$\bullet (\forall x \exists y): y+1 = x$
Negates to
$(\exists x \forall y): y+1 \not= x$

The first statement is False for the set of Natural Numbers (for example), but True for the Integers (for example). Obviously one “swaps” these “Truth Values” for the second statement (by the “double negation law”!… ain’t logic fun?).

• ## (Partial) Contents Page

Vlorbik On Math Ed ('07—'09)
(a good place to start!)