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I have this task in mathematical logic for which I don't really have a tool for solving.

What statement is true for every a, b and c?

  1. $a \in b \wedge b \in c \rightarrow a \in c$

  2. $a \in b \wedge b \subseteq c \rightarrow a \in c$

  3. $a \subseteq b \wedge b \in c \rightarrow a \in c$

  4. $a \subseteq b \wedge b \subseteq c \rightarrow a \subseteq c$

I need to determine which of these is true and need to explain my answer for every single one of these. Now, intuitively, all of these would be true (1. is trivial, 2. if a from b, b is a subset of c and then a in c is true because a is in b and b is a subset of c, 3. a subset of b, b from c then a is from c as well also should make sense (unless here $\in$ is meant to represent at least one element like for example theres element from b in c but doesnt necessarily need to include whole area of b? I consider $\in$ to represent an entire set belonging in another set, basically like a strict inclusion)).

I think my intuition is off and I'm definitely missing a tool required to solve this. What tools are required for this and how to, after solving, explain my reasonings?

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  • $\begingroup$ $a \in b$ means "$a$ is a member of $b$" and is not some kind of inclusion. Statements (1) and (3) are wrong. $\endgroup$
    – Jean Abou Samra
    Commented Jun 13 at 18:06
  • $\begingroup$ @JeanAbouSamra Thank you for a quick response, I'm not just looking for answers but rather an explanation on how to get them. Okay so $a$ being a member of $b$ and $b$ being a member of $c$ would mean what exactly? Does that mean that $a$ is just one element from $b$ and then one element from $b$ is an element of $c$ therefore not necessarily meaning $a \in c$? $\endgroup$
    – Danilo Jonić
    Commented Jun 13 at 18:17
  • $\begingroup$ I'm not sure I understand what you mean exactly. For example, $0 \in \{0\}$, $\{0\} \in \{\{0\}\}$, but $0 \not\in \{\{0\}\}$. $\endgroup$
    – Jean Abou Samra
    Commented Jun 13 at 18:21
  • $\begingroup$ @JeanAbouSamra I understand your example as it's direct application of the properties of a partitioned set. But, I found axiom of a partition set stating that $x \in P(a) \Leftrightarrow x \subseteq a$? So for your example, taking $x=0$, it obviously is true that $0 \in P(0)$ but it would also be true that $0 \subseteq 0$ so it's then trivial. I think I understand it now. $\endgroup$
    – Danilo Jonić
    Commented Jun 13 at 18:39
  • $\begingroup$ @DaniloJonić Hey, thanks for editing away my typo below. Not important, but I might as well explain: I rejected changing #4's 'that is' to 'so', because while the word 'so' definitely means ⟹, I'm deliberately choosing the phrase 'that is' to suggest specifically the definition the Subset relation is transitive ⇔ ∀a,b,c (a⊆b∧b⊆c → a⊆c). $\endgroup$
    – ryang
    Commented Jun 14 at 3:33

1 Answer 1

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  1. $a \in b \wedge b \in c \rightarrow a \in c$

Counterexample (the antecedent is satisfied but not the consequent): $$a=1,\;b=\{1\},\;c=\big\{\{1\}\big\}.$$

  1. $a \in b \wedge b \subseteq c \rightarrow a \in c$

Consider any $a,b$ and $c$ and suppose that $a$ is an element of $b$ and $b$ is a subset of $c.$ So, $c$ contains every element of $b,$ including $a.$ So, $a$ is an element of $c.$ Hence, the given formula is universally true.

  1. $a \subseteq b \wedge b \in c \rightarrow a \in c$

Counterexample: $$a=\{1\},\;b=\{1,2\},\;c=\big\{\{1,2\}\big\}.$$

  1. $a \subseteq b \wedge b \subseteq c \rightarrow a \subseteq c$

The 'subset' relation is transitive; that is, the given formula is universally true.

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