Proving simple sequence using natural deduction

Question

I need to prove below reasoning using natural deduction: Santa always wears either shirt or t-shirt but never both at the same time. He never has a jacket and bow tie at the same time. If he has a t-shirt he always takes his jacket. Then we can conclude that if Santa has a bow tie, he also has a shirt.

k : Santa has a jacket

s : Santa has a shirt

f : Santa has a bow tie

t : Santa has a t-shirt

This is what I have done so far:

s → ¬t, k→ ¬f, t → k   ⊢   f→ s ¨

1   s → ¬t  premis
2   k→ ¬f   premis
3   t → k   premis
------------------------------proof-box
4   f       assumption
5   ¬k      MT 2,4
6   ¬t      MT 3,5
------------------------------end of proof-box
7   f →¬t   →i 4-6

But how can i get from ¬t to s using natural deduction rules?

Answer 1

Here is a proof via natural deduction:

1. $(s \vee t) \wedge \neg(s \wedge t)$ premise
2. $\neg (k \wedge f)$ premise

3. $t \rightarrow k$ premise

   ---

4. $\neg k \vee \neg f$ DeMorgan's Rule 2

5. $k \rightarrow \neg f$ implication 4
6. $t \rightarrow \neg f$ hypothetical syllogism 3,5
7. $s \vee t$ conjunction elimination (aka simplification) 1
8. $\neg \neg s \vee t$ double negation 7
9. $\neg s \rightarrow t$ implication 8
10. $\neg s \rightarrow \neg f$ hypothetical syllogism 6,9
11. $f \rightarrow s$ contrapositive law 10

$\therefore f \rightarrow s$

You were on the right track, you just needed $s \vee t$ to deduce your conclusion. Remember that conjunction elimination permits you to split propositions joined by conjunctions onto new lines.