The goal to be proved here is Theorem 2.86 from the famous *Principia Mathematica*
of 1910 (Vol 1), written by Whitehead and Russell. The goal is a conditional:
$$[(\phi \rightarrow \psi) \rightarrow (\phi \rightarrow \delta)] \rightarrow
\phi \rightarrow (\psi \rightarrow \delta).$$
No remaining
use of an oracle can be present in a proof that constitutes a solution to this problem,
but as always, use of the oracle along the way is not only permitted, but encouraged.

Solve

**Deadline ** 28 Feb 2019 11:45:00 EST

Solve

The theorem to be proved here is a straightforward conditional: viz.
$$\vdash (A \rightarrow B) \rightarrow ((B \rightarrow C) \rightarrow (A \rightarrow C)).$$
No remaining
use of an oracle can be present in a proof that constitutes a solution to this problem.

Solve

**Deadline ** 22 Feb 2019 12:00:00 EST

Solve

This problem, known as ‘Bringsjord’ 1 in the initial, motivating part of the of the Spring 2019 edition of the class, has been shown to be unsolvable by the vast majority of college students who haven‘t had sustained, effective training in formal logic.

The problem presents two givens:

- The following three propositions are either all true, or all false.
- If Billy helped, Doreen helped.
- If Doreen helped, Frank helped.
- If Frank helped, Emma helped.

- Billy helped.

And the question is: Does it follow deductively from these two givens that Emma helped? It seems that the minds of those pondering this question can be partitioned into three categories: The first category is composed of those who answer in the affirmative, because they simply focus on chaining through the conditionals to arrive at the conclusion that Emma helped. The second category is composed of those who realize those in the first category have failed to take account of the all-false possibility, and are quite sure that when the three conditionals are all false, it can’t be deduced that Emma helped. The third category is made up of those who are truly logical; they realize that even in the case when all three conditionals are false, it can be proved that Emma helped. The third group is correct, and hence the answer to the question is: Yes!

While as is standard you can use the PC provability oracle *while* working toward your
solution, no remaining use of the oracle is permitted in the submitted file, if a trophy is
to be won.

View

The goal to be proved here is Frege's axiom THEN-2 his axiomatization of the
propositional calculus. No remaining use of any oracle can be present in the
final solution.

View

**Deadline ** 14 Mar 2019 12:45:00 EST

View

In this problem you are provided with three biconditionals, and must deduce a new
one based on the endpoints of what can be regarded as the chain of the trio.
One might, upon saving the maneuver in one‘s mind or one‘s library,
dub the generalization of
what a successful proof shows as *biconditional intro by chaining* (although you
will no doubt be able to devise a less verbose label for the tactic in question).
Expressed metalogically, your overall task is to validate:
$$\{P \leftrightarrow Q, Q \leftrightarrow R, R \leftrightarrow S\} \vdash P \leftrightarrow S.$$
You are of course free to use the PC oracle *during* the construction of your proof, but no
use of the oracle can remain in your finished, for-credit proof.
(No deadline for now.)

View

View

In this challenge you are given a peculiar biconditional, and asked to prove from it that
the moon is made of green cheese (symbolized simply as $G$, as is our custom).
It’s possible to obtain
$G$ from the biconditional alone, but to make the problem
easier, a second given, The Law of the Excluded
Middle (LEM), or by its Latin name *tertium non datur*,
is provided. We encourage ambitious students to leave aside this
second given and, if they wish to use LEM, to prove it as a theorem
(or more precisely, a lemma) in the course of reaching $G$.
So, minimally you need to prove this: $\{ P \leftrightarrow \neg P, P \vee \neg P\} \vdash G$;
but the more impressive proof would establish this:
$\{ P \leftrightarrow \neg P\} \vdash G$.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

As the saying goes, cheaters (ultimately) never prosper.
In the particular case at hand, a formalization of this
saying is to be proved from a single given, to wit:
$$\forall s \forall t [\mathit{Cheats}(s, \mathit{reddit}, t) \rightarrow
\forall t' (t < t' \rightarrow \neg \mathit{Prospers}(s, t'))].$$
The finished proof *can* have in it remaining use of the PC provability oracle,
but no use of any other
oracle can be in the finished proof.
Notice that in the pre-set file that opens in HyperSlate^{TM}, an English version of the given
formula is included.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

Here you are simply asked to prove that the *contrapositive* of a conditional
can be deduced from that conditional. This is expressed in meta-logic by this
statement: $\{ \phi \rightarrow \psi \} \vdash \neg \psi \rightarrow \neg \phi$. The problem
NYS 2 tested for an understanding of this statement, which when thought of as an inference
rule or schema is referred to as *contraposition* or *transposition*.
No use of an oracle can remain in your solution.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

Here you are tasked with proving the very
handy *hypothetical syllogism*,
expressed in meta-logic by this
statement:
$\{ \phi \rightarrow \psi, \psi \rightarrow \delta \} \vdash \phi \rightarrow \delta$.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

The theorem to be proved here says that if Kris speaks Dutch, Santa Claus exists.
The English sentence “Santa Claus exists” is symbolized here
as $\exists x (x = \mathit{santa})$, i.e. as “There exists an *x* such
that it’s Santa.”
Cinematic cognoscenti will not only realize that this problem is derived from
*Miracle on 34th Street*; they will also peg the specific scene in the
movie that is being alluded to. No remaining use of an oracle is permitted in
this problem.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

The challenge here is to prove that from Russell's instantiation of Frege's
doomed Axiom V a contradiction can be promptly derived. The letter has of course
been examined in class, in some detail, by S Bringsjord; it, along with an astoundingly soft-spoken
reply from Frege, can be found
here.
Put meta-logically, your task in the present problem is to build a proof that
confirms this:
$$\{ \exists x \forall y ((y \in x) \rightarrow (y \not\in y)) \} \vdash \zeta \wedge \neg \zeta.$$
Make sure you understand that the given here is an instantiation of Frege’s Axiom V; i.e.
it’s an instantiation of $\exists x \forall y ((y \in x) \rightarrow \phi(y))$.
**Note**: Your finished proof is allowed to make
use the PC provability oracle (but *only* that oracle).
(No deadline for now.)

Solve

Solve

This straightforward problem is quickly solved with a minimum of tedium, courtesy of
the PC (entailment) provability oracle, use of which is allowed to remain in your finished proof
(but no use of any other oracle can be in the finished proof).
(No deadline for now.)

Solve

Solve

Here your task is to prove the explosion inference schema/rule
in HyperSlate^{TM}; the schema says that from a contradiction
anything whatsoever follows. Expressed in meta-logic, you are to prove
$\{\phi \wedge
\neg \phi\} \vdash \psi$. As explained in class, for whatever reason, while
the general proof technique of *reductio ad absurdum*
(= proof by contradiction, indirect proof, etc.) is covered in high-school math
(at least it is in the textbooks, e.g. solid ones for Geometry and Algebra 2), explosion
for some reason is almost invariably left aside, despite the indisputable
dual fact that explosion is valid
in classical logic and mathematics, and that it can be extremely useful.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

This is a straightforward problem in which you are to create a HyperSlate^{TM} proof that
confirms this:
$$\left\{ \begin{aligned}&\neg \exists x (\texttt{Small}(x) \wedge \texttt{Llama}(x)),\\
&\forall x (\texttt{Small}(x) \vee \texttt{Medium}(x) \vee \texttt{Large}(x)) \end{aligned}\right\} \\
\vdash \forall x \Big(\texttt{Llama}(x) \rightarrow \big(\texttt{Medium}(x) \vee \texttt{Large}(x)\big)\Big)$$
**Note**: Your finished proof is allowed to make use of the PC provabiity oracle,
but of no other oracle.
(No deadline for now.)

Solve

Solve

Is the Moon made of green cheese? Apparently, if one denies what would appear to be a rather
crazy claim (viz. that if some proposition $P$ holds, every proposition implies $P$), it does
indeed follow that the Moon is made of cheese. Your task here is to build a proof that confirms
this (i.e. a proof that confirms $\{ \neg (P \rightarrow (Q \rightarrow P))\} \vdash G$), where we
take $G$ to symbolize that the Moon is made of green cheese.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

This problem requires that one employ the inference schema (or rule) known as
*disjunction elimination*, a.k.a. *proof by cases*.
The inference schema, the hypergraphical specification of which
can be found in the chapter **Propositional Calculus** in LAMA-BDLA,
is based on the core idea that if one has a number of cases/disjuncts
$\phi_1 \vee \phi_2 \vee \ldots \vee \phi_k$ to consider, and can show that
$\psi$ follows from each $\phi_i$ (made as an assumption), then $\psi$ follows
from the disjunction itself.
(No deadline for now.)

Solve

Solve

Your challenge here is to shine the light of logic on one of the most famous cognitive
illusions created by the august cognitive scientist (and early AI pioneer!)
P. Johnson-Laird, and to thereby dissolve the illusion, which can
be presented as follows:
*can* correctly deduce is that there *isn't* an ace
in the hand!
Expressed in meta-logic, the HyperSlate^{TM} file in question challenges
you to produce a proof that confirms:
$$\left\{ \begin{aligned}( K \rightarrow A & \vee \neg K \rightarrow A) \wedge \\
\neg (K \rightarrow A &\wedge \neg K \rightarrow A) \end{aligned}\right\} \\ \vdash \neg A$$ One
of the cool things
about this problem is that DeMorgan's Law is key, and while that's something many students of
high-school math will be acquainted with, few will unfortunately have been asked to prove it.
(Having not been asked is unfortunate because unless one can prove such a thing, it's rather
doubtful that one understands it.)

Solve

If there is a king in the hand, then there's an ace in the hand; or if there isn't a king in the hand, then there's an ace in the hand; not both of these if-then statements are true. What can you correctly deduce?Even professors across the globe (as long as they aren't motivated logicians or mathematicians) join 99% of the general public in declaring: 'That there's an ace in the hand!' Unfortunately, as you now know from Bringsjord's relevant lectures (the full lineup of which can be found here), not only is this wrong, but what you

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

The challenge here is to establish the so-called *Barber Paradox*.
You are to show that from the asertion that in a certain town there resides a
(male) barber who shaves all and only the men of that town who don't shave themselves,
a contradiction can be deduced. (This is of course a picturesque version of Russell's
Paradox.)
The preset HyperSlate^{TM} file conveniently supplies a first-order symbolization of
the paradoxical proposition.
You are premitted to leave use of the PC provability oracle in your finished proof
(but no other oracles can remain).

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

Here one needs to provide a proof that confirms
$\{\phi \wedge \psi\} \vdash \psi \wedge \phi$.
Once the proof is found, this shows that the order of conjuncts is
immaterial in logical consequence, for clearly the number of conjuncts could be expanded arbitrarily,
so we declare that once this proof works, it's mere tedium to handle $\phi_1 \wedge \phi_2 \wedge \ldots
\wedge \phi_k,$ for any $k \in \mathbb{N}$.

Solve

**Deadline ** 21 Mar 2019 01:57:00 EDT

Solve

This warmup for the infamous problems that typically follow (in Bringsjord’s demonstration
that there’s a dire need to learn formal logic) was covered in the first of the two “motivating”
classes, and is taken from New York State's high-school math coverage (under algebra).
Viewed from the perspective of meta-logic, your task is to prove
$\{\neg A \vee \neg B, B, C \rightarrow A\} \vdash \neg C$.
Here, $\neg C$ is the key among a
series of options; the distractors are $C, \neg B, H, A$ and NONE OF THE ABOVE.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

Viewed from the perspective of meta-logic, your task is to prove
$\{\neg \neg C, C \rightarrow A, \neg A \vee B, B \rightarrow D, \neg (D \vee E)\} \vdash H$. Since
of course $H$ here is arbitrary, and doesn't appear in the givens,
what your proof will demonstrate, if valid, is that from the set
of given formulae in the propositional calculus, *any* $\phi$ can be derived.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve

The theorem to be proved here is *tertium non datur*, a.k.a. The Law of the Excluded Middle;
you will need to prove this: $\vdash \phi \vee \neg \phi$.
For some edifying supplementary reading, provided for the motivated, consult the SEP entry on
Contradiction.

Solve

**Deadline ** 30 Apr 2019 23:59:00 EST

Solve