2 column proof - m ^ [(m^s)->r]^~r=>~s A whale is a mammal (m). If a whale is a mammal and a spider has six legs (s), then a frog is a reptile (r).  But, a frog is not a reptile.  Therefore, a spider does not have six legs.
Work:
m = a whale is a mammal
s = a spider has six legs
r = a frog is a reptile

P1: m
p2: (m ^ s) -> r
p3: ~r
therefore:  ~s

Steps                 Reason
1.  m                 p1 - given
2.  (m ^ s) -> r      p2 - given
3.  ~r                p3 - given
4.  ???

Apologies. I'm trying to help my daughter with this and I don't know the next step.  We are trying to do the conjunction of any 2 of 1-3, but that doesn't result in any of the theorems that her cheat sheet shows.
 A: If you want to prove that $s$ is false in some context, then you can (in that context) ask whether it is possible that $s$ is true or not. If there is no way that $s$ can be true in that context, then it must be that $s$ is false. (Related to this is proof by contradiction, in that the underlying reasoning is the same.) For concreteness, here is how the logical reasoning to answer your question would go:
$\def\imp{\rightarrow}$
$m$.   [assumption]
$( m \land s ) \imp r$.   [assumption]
$\neg r$.   [assumption]
If $s$ then:
  $m$.   [Something that is true is still true in more restricted conditions (here under "$s$ is true").]
  $s$.   [We are working in the restricted context of "$s$ is true".]
  $m \land s$.   [Therefore in this context $m \land s$ is true.]
  $( m \land s ) \imp r$.   [As before, we're restating something already true outside the current context.]
  $r$.   [By meaning of $\imp$.]
  $\neg r$.   [Another restatement.]
  Contradiction.   [This merely means that the current context is impossible.]
Therefore $\neg s$.   [The only possibility left is the opposite of the impossible context.]
