Symbolize and provide
derivations for each of the following.
1. Every student is
ambitious. No one who is ambitious will be lazy. Therefore,
there will be no lazy students.
(x)(Sx ->
Ax), (x)(Ax -> ~Lx) |- (x)(Sx -> ~Lx)
1.
(x)(Sx -> Ax)
2.
(x)(Ax -> ~Lx) \show (x)(Sx
-> ~Lx)
3.
Sa ->
Aa
1, UI
4.
Aa ->
~La
2, UI
5.
Sa -> ~La
3, HS
6.
(x)(Sx -> ~Lx) 5, UG
2. Some students
do work hard. Anyone who works hard will pass.
Therefore, there are students who will pass.
(Ex)(Sx
& Wx), (x)(Wx -> Px) |- (Ex)(Sx & Px)
1.
(Ex)(Sx & Wx)
2.
(x)(Wx -> Px)
\ show (Ex)(Sx & Px)
3.
Sa &
Wa
1, EI
4.
Wa ->
Pa
2, UI
5.
Wa
3, Simp
6.
Pa
4,5, MP
7.
Sa
3, Simp
8.
Sa &
Pa
6,7, Conj
9.
(Ex)(Sx & Px) 8, EG
3. Alice is an
ambitious student
and Bob is a bright student. If there are both bright students
and
ambitious students, then the teachers will be happy. No teachers,
then, will be unhappy.
Aa & Sa,
Bb & Sb, [(Ex)[(Bx & Sx) & (Ex)(Ax & Sx)] -> (x)(Tx
-> Hx) |- (x)(Tx -> Hx)
1.
Aa & Sa
2.
Bb & Sb
3.
[(Ex)(Bx & Sx) & (Ex)(Ax & Sx)] -> (x)(Tx
-> Hx) \ show (x)(Tx -> Hx)
Note
that in the premise we are not saying there is an individual who is
both ambitious and bright, so we need to use separate
quantifiers. Also note that in "translating" the conclusion we
are already building in the idea of a double negative.
4
(Ex)(Bx &
Sx)
2, EG
5.
(Ex)(Ax &
Sx)
1, EG
6.
(Ex)(Bx & Sx) & (Ex)(Ax &
Sx)
4,5, Conj
7.
(x)(Tx ->
Hx)
3,6, MP
4. Some classes are not
interesting. Any class that is not interesting will not be
fun. If
there are classes that are not fun then Bob will not be happy. It
does
follow that Bob will be unhappy.
(Ex)(Cx
& ~Ix), (x)[(Cx & ~Ix) -> ~Fx], (Ex)(Cx & ~Fx) ->
~Hb |- ~Hb
1.
(Ex)(Cx & ~Ix)
2.
(x)[(Cx & ~Ix) -> ~Fx]
3.
(Ex)(Cx & ~Fx) -> ~Hb / show ~Hb
Note the importance of
grouping correctly (using our parentheses and brackets as we
should). In the second premise the quantifier is the main
connective, but in the third premise it is the arrow.
4.
Ca & ~Ia
1, EI we are supposing class "a" (algebra, for
instance) as our example but we are not really claiming that it would
be class "a"
5.
(Ca & ~Ia) ->
~Fa
2, UI if
the uninteresting class really is class "a" we would know it is not fun
6.
~Fa
4,5, MP
7.
Ca
4, Simp
8.
Ca &
~Fa
6,7, Conj
9.
(Ex)(Cx &
~Fx)
8, EG
10.
~Hb
3,9, MP
5. Alice will pass only if all the classes are
interesting. Only classes that are fun are
interesting. Some classes are not fun, so it follows that Alice
is not going to pass.
Pa ->
(x)(Cx -> Ix), (x)(Ix -> Fx), (Ex)(Cx & ~Fx) |- ~Pa
Note the importance of
symbolizing in the right order when the word "only" indicates a
necessary condition.
1.
Pa -> (x)(Cx -> Ix)
2.
(x)(Ix -> Fx)
3.
(Ex)(Cx & ~Fx) / show ~Pa
4.
Ca &
~Fa
3, EI
5.
Ia -> ~Fa
2, UI
6.
~Fa
4, Simp
7.
~Ia
5,6, MT
8.
Ca
4, Simp
9.
Ca & ~Ia
7,8, Conj
10.
(Ex)(Cx & ~Ix) 9, EG
remember, even if we wanted to, we could
not now come up with a universal generalization based on the name "a"
11.
~(x)(Cx -> Ix)
10, QN
12.
~Pa
1,11, MT
