MHB Adjacency Matrix Problem and Alphabet Problem

  • Thread starter Thread starter sadsadsadsa
  • Start date Start date
  • Tags Tags
    Matrix
sadsadsadsa
Messages
15
Reaction score
0
Could some please tell me if they think my answer for 1c and 3d of these questions I've done are right. thanks.View attachment 4393View attachment 4394
 

Attachments

  • 1212-small.jpg
    1212-small.jpg
    41.4 KB · Views: 101
  • 11113-small.jpg
    11113-small.jpg
    113.7 KB · Views: 114
Physics news on Phys.org
These appear to be for credit. If so, it is against forum policy knowingly to help with such problems. Please PM if it is otherwise. Thread closed.
 
Hello, I'm joining this forum to ask two questions which have nagged me for some time. They both are presumed obvious, yet don't make sense to me. Nobody will explain their positions, which is...uh...aka science. I also have a thread for the other question. But this one involves probability, known as the Monty Hall Problem. Please see any number of YouTube videos on this for an explanation, I'll leave it to them to explain it. I question the predicate of all those who answer this...
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top