Let ≿ be a complete preference relation (as in game theory). How to prove that ≿ is transitive if and only if ≻ and ∼ are both transitive?
My reasoning is as follows. a ≿ b probably means (a ≻ b) ∨ (a ∼ b), by analogy with ≥, which means greater OR equal. Transitivity means (a ≿ b) ∧ (b ≿ c) → (a ≿ c). Therefore we must prove that following expression is valid:
$$[(a \succ b) \wedge (b \succ c) \rightarrow (a \succ c)] \wedge [(a \sim b) \wedge (b \sim c) \rightarrow (a \sim c)] \equiv [((a \succ b) \vee (a \sim b)) \wedge ((b \succ c) \vee (b \sim c)) \rightarrow ((a \succ c) \vee (a \sim c))]$$
For brevity, let's substitute the relations with capital letters:
a ≻ b: A
a ∼ b: B
b ≻ c: C
b ∼ c: D
a ≻ c: E
a ∼ c: F
The expression then becomes:
$$(A \wedge C \rightarrow E) \wedge (B \wedge D \rightarrow F) \equiv [(A \vee B) \wedge (C \vee D) \rightarrow (E \vee F)]$$
No amount of logical manipulation allows to prove that LHS is equivalent to RHS. What am I doing wrong? Ideally, your answer would consist of a hint that would allow me to solve the problem myself without giving out the answer.
I would start with the most complex side of the statement you want to prove, which in this case is the left hand side, expand your definitions, and take it from there.
$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\rightarrow} \newcommand{\followsfrom}{\leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} \newcommand{\Rg}{\succ} \newcommand{\Re}{\sim} \newcommand{\Rge}{\succsim} $(As a matter of notation, I'm leaving out the parentheses around $\;a \Rge b\;$ etc., to make things more readable.)
So we calculate as follows, starting with the left hand side: $$\calc a \Rge b \;\land\; b \Rge c \op\equiv\hint{definition of $\;\Rge\;$} (a \Rg b \lor a \Re b) \;\land\; (b \Rg c \lor b \Re c) \op\equiv\hints{logic: distribute $\;\land\;$ over $\;\lor\;$, three times} \hints{-- we want to use transitivity of $\;\Rg\;$ and $\;\Re\;$, and this is} \hint{the simplest way to bring both $\;\Rg\;$ and $\;\Re\;$ together} (a \Rg b \land b \Rg c) \;\lor\; (a \Rg b \land b \Re c) \;\lor\; (a \Re b \land b \Rg c) \;\lor\; (a \Re b \land b \Re c) \op\then\hint{...} \endcalc$$ Now you should be able to finish this calculation. (Note that you will be using the law of logic which says that if $\;P \then Q\;$, then also $\;P \lor R \then Q \lor R\;$.)