Lecture 4 - Predicates & Quantifiers (Contd.)
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange askers cite it when a worked example or claim from the instructor confuses them, such as quantifier equivalences and translating English statements into first order logic.
Recommended by 1 people across 1 different months, 2015 to 2020.
What it is
A lecture from an IIT discrete mathematics course on NPTEL, continuing the treatment of predicates and quantifiers in first order logic. It covers translating statements into predicate logic and equivalences between quantified formulas.
What people said
Recommends Chris Bedford · 2015-06 · Mathematics Stack Exchange · on “confusion about 2 first order logic wff's - they seem not equal, but instructor says they are =”
I had a question about two first order logic formulas given in this lecture [this video] in the series on Discrete Mathematical Structures from IIT. The instructor says (at 36:19 in the video) that the following two statements are equivalent: $\exists x (P(x) => Q(x) )$ $\forall x P(x)$ => $\exists x Q(x)$ In the first statement my understanding is that X is bound such that we are talking about the same value of x for P(x) and Q(x). In words, statement 1 seems to be saying that if we have P(3), then that implies Q(3)... not Q(4) or Q(5) or some other number. In the second statement it seems there are two bindings of 'x'. For any 'x' (say x=3) P(x) (that is, P(3) in this case) implies that there e
Cites srt1104 · 2017-05 · Mathematics Stack Exchange · on “I am getting two answer for this 1st order logic question.”
I had a doubt about a first order logic question given in this lecture [this video] which is a series on Discrete Mathematical structures from IIT. If $x*y=x$ for all $y$ , then $x=0$ . You have to represent this in predicate logic. The universe of discourse is set of non-negative integers. Let $P(x,y,z)$ represent: $x*y=z$ Answer given by the instructor is; $$∀x[∀y P(x,y,x) => x=0]$$ Which I understood and is true. But the answer I managed was: $$∃x[∀y P(x,y,x) ∧ x=0]$$ This also seems right. Also if you expand the first assertion, $$∀x[∀y P(x,y,x) => x=0]$$ $$∀x[¬∀y P(x,y,x) ∨ x=0]$$ $$∀x ¬[∀y P(x,y,x) ∧ x≠0]$$ $$¬∃x[∀y P(x,y,x) ∧ x≠0]$$ Which is not the same as I managed but I am not able to f
Cites SKUND VERMA · 2020-07 · Mathematics Stack Exchange · on “Can't we say that "some numbers are not real" is the negation of "all numbers are real"?”
Refer to this video [this video] On 52:49, can't we say that "some numbers are not real" is the negation of "all numbers are real"?