Argue that the relation "logically implies" is transitive on the set of all statements.
I've never seen an argument for transitivity, my textbook only gives examples.
Hmm, does this mean that your textbooks only gives examples of transitive relations, without proofs that they are actually transitive? If there are proofs or explanations, then those are arguments... Does your textbook have the definition of transitivity?
A relation $\displaystyle R\subseteq A\times A$ on a set $\displaystyle A$ is called transitive if for all $\displaystyle x, y, z\in A$, if $\displaystyle (x,y)\in R$ and $\displaystyle (y,z)\in R$, then $\displaystyle (x,z)\in R$.
To show that a relation is transitive you need to thoroughly understand this definition, and you need to know how to prove statements of the form "for all x, ..." and "if ..., then ...".
the transitivity of (logical) implication, then, would mean:
$\displaystyle ((p \implies q)\wedge(q\implies r)) \implies (p \implies r)$
is this a true statement for "$\displaystyle \implies$" (logically implies)?
(hint: assign values of true and false for p,q and r (you will get 8 possible combinations you have to try), and examine the truth-values of
$\displaystyle ((p \implies q)\wedge(q\implies r))$ and $\displaystyle (p \implies r)$
you can use the following tables:
A B | A→B (A implies B)
T T | T
T F | F
F T | T
F F | T,
A B | A ∧ B (A and B)
T T | T
T F | F
F T | F
F F | F.....what are your findings?)
the two-line arrow is technically "correct" (means "implies") as the one-line arrow usually means "goes to" (used for limits and mappings).
(however, looking at a POset as a category, they are the same thing: as implication provides a partial ordering on wffs).