Symbolize the following valid
arguments and provide derivations using
the full package of equivalence and inference rules that we now have.
1. If logic is fun and interesting, then it should be easy.
Logic
turns out to be interesting even though it is not easy, so it must not
be fun.
(F & I) -> E, I &
~E |- ~F
1. (F & I) -> E
2.
I &
~E
/show ~F
3.
~E
2, Simp
4.
~(F &
I)
1,3, MT
5.
~F v
~I
4, DM
6.
I
2, Simp
7.
~F
5,6, DS
2. Logic is easy if it is fun, and it is fun if it is
interesting.
Alice will study unless logic is not easy, but since she is not
studying it must not be interesting.
F -> E, I -> F, S v ~E, ~S
|- ~I
1. F -> E
2.
I -> F note that it makes things easier to
express this as a separate premise right away
3.
S v ~E
4.
~S
/show ~I
5.
I -> E
1,2, HS
6.
~E 3,4, DS
7. ~I
5,6, MT
3. Only if logic is not either
easy or fun will it not be interesting.
Unless logic is easy Alice will not study. Alice is studying,
so this proves logic must be interesting.
~I -> ~(E v F), E v ~S, S |- I
1.
~I -> ~(E v F)
2.
E v ~S
3.
S
/ show I
4.
E
2,3, DS
5.
(E v F) -> I 1, Trans
6.
E v
F
4, Add
7.
I
5,6, MP
4. Alice will study only
if logic is both easy and fun. Also, Alice will
pass only if she studies. It follows that Alice will pass only if
logic is easy.
S -> (E & F), P -> S |- P
-> E
1. S -> (E & F)
2.
P ->
S
/ show P -> E
3.
P -> (E & F)
1,2, HS
4.
~P v (E &
F)
3, Impl
5.
(~P v E) & (~P v F)
4, Dist
6.
~P v
E
5, Simp
7.
P ->
E
6, Impl
5. Every employee needs
to
work hard if there are to be some companies that are successful.
Some
employees are not working hard. That means no companies are
successful.
(Ex)(Cx & Sx) -> (x)(Ex -> Wx),
(Ex)(Ex & ~Wx) |- (x)(Cx -> ~Sx)
1.
(Ex)(Cx & Sx) -> (x)(Ex ->Wx)
2.
(Ex)(Ex &
~Wx)
/ show (x)(Cx -> ~Sx)
3.
~(x)(Ex ->
Wx)
2, QN
4.
~(Ex)(Cx &
Sx)
1,3, MT
5.
(x)(Cx ->
~Sx)
4, QN
