Q:

Proving simple sequence using natural deduction

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?

A:

Here is a proof via natural deduction:

$(s \vee t) \wedge \neg(s \wedge t)$ premise
$\neg (k \wedge f)$ premise
$t \rightarrow k$ premise

$\neg k \vee \neg f$ DeMorgan's Rule 2
$k \rightarrow \neg f$ implication 4
$t \rightarrow \neg f$ hypothetical syllogism 3,5
$s \vee t$ conjunction elimination (aka simplification) 1
$\neg \neg s \vee t$ double negation 7
$\neg s \rightarrow t$ implication 8
$\neg s \rightarrow \neg f$ hypothetical syllogism 6,9
$f \rightarrow s$ 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.

