Tag Archives: truth tables

Logical Dead End

One is reduced to hysterical laughter to try to maintain a modicum of sanity.

Our Senate at work: Republican Mitch McConnell said (Dec 6) “Legislation that doesn’t include policy changes to secure our borders will not pass the Senate.”  Republican Trump said (Feb 3) the Senate should not pass legislation that includes border security.  Let P be the statement “Senate legislation should include border security.” and let Q be the statement “Senate should pass legislation.”  Then we have the Republicans saying

(~P ⇒ ~Q) ˄ (P ⇒ ~Q)

Show that this is equivalent to ~Q, that is, “The Senate should not pass legislation.”—basically stop working.

It looks like the Republicans in the House are doing the same thing:

politico.com

See Logical Dead End

Old Codger Rant, with Update (4/24/2024):  Continue reading →

Peirce’s Law

The June 2023 Carnival of Mathematics # 216 at Eddie’s Math and Calculator Blog has the rather arresting item concerning Peirce’s Law from the American logician Charles Sanders Peirce (1839 – 1914).

“Peirce’s Law:  Jon Awbrey of the Inquiry Into Inquiry blog

This article explains Pierce’s Law and provides the proof of the law.  The proof is provided in two ways:  by reason and graphically.  Simply put, for propositions P and Q, the law states:

P must be true if there exists Q such that the statement “if P then Q” is true.  In symbols:

(( P ⇒ Q) ⇒ P) ⇒ P

The law is an interesting tongue twister to say the least.”

Perhaps another way of saying it is “if the implication P ⇒ Q implies that P is true, then P must be true.”  Still, it sounds weird.

See Peirce’s Law

(Update 6/20/2023)  Appendix: Valid Argument Continue reading →

“Fermat’s Last Theorem” Puzzle

Here is a mind-numbing logic puzzle from Futility Closet.

“A puzzle by H.A. Thurston, from the April 1947 issue of Eureka, the journal of recreational mathematics published at Cambridge University:

Five people make the following statements:—

Which of these statements are true and which false?  It will be found on trial that there is only one possibility.  Thus, prove or disprove Fermat’s last theorem.”

Normally I would forgo something this complicated, but I thought I would give it a try.  I was surprised that I was able to solve it, though it took some tedious work.  (Hint: truth tables.  See the “Pointing Fingers” post regarding truth tables.)

One important note.  The author is a bit cavalier about the use of “Either …, or …”.  In common parlance this means “either P is true or Q is true, but not both” (exclusive “or”: XOR), whereas in logic “or” means “either P is true or Q is true, or possibly both” (inclusive “or”: OR).  I assumed all “Either …, or …” and “or” expressions were the logical inclusive “or”, which turned out to be the case.

See the Fermat’s Last Theorem Puzzle

Pinocchio’s Hats

This problem in logic from Presh Talwalkar recalled an article I wrote a while ago but did not publish.  So I thought I would post it as part of the solution.

“Assume that both of the following sentences are true:

  1. Pinocchio always lies;
  2. Pinocchio says, “All my hats are green.”

We can conclude from these two sentences that:

  • (A) Pinocchio has at least one hat.
  • (B) Pinocchio has only one green hat.
  • (C) Pinocchio has no hats.
  • (D) Pinocchio has at least one green hat.
  • (E) Pinocchio has no green hats.”

Actually, the question is which, none or more, of statements (A) – (E) follow from the two sentences?

Answer.

See Pinocchio’s Hats for solutions.