In the first lesson we previewed the idea
of a derivation. I'll repeat this below:
Now, let's bring in two rules. One allows us to eliminate the C, the other to exchange one string or part of a string for another.
This is the first. If we have any conditional
and we have the part to the left (the antecedent) by itself, we can also
have the part to the right (the consequent) by itself. We represent
the rule in a general manner this way:
PQC, P |- Q
C elim (the sign |- represents that whenever we
have what is the left the thing to the right will follow)
This is the second. We can switch the
antecedent and the consequent if we negate both.
PQC = QNPNC
Contra (for contraposition)
What we will be working with very soon are
derivations, not too different from what you saw in the letter game.
Let's look at a sample:
PQC, QN |- PN
1. PQC Ass
2. QN Ass
3. QNPNC
1 Contra
4. PN
2,3 C elim
If you understand the principle here, you will
also see how we can work with any substitution instances. Let's say
the goal is to show this: PQARSOC, RSON |- PQAN
Now watch how we go through the same set of
moves.
1. PQARSOC Ass
2. RSON
Ass
3. RSONPQANC 1 Contra
4. PQAN
2,3 C elim
Where we go next is in working with some additional rules that will let us show more complicated relationships.
Let's add these equivalence rules (meaning,
we have patterns in which the truth tables will be identical for both expressions):
PQC = PNQO
MI ( for "material implication") in
SLIPS I label this CO subs
PNN = P
DN (double negation) in SLIPS I label
this N subs
PQAN = PNQNO DeM
(DeMorgan's laws) in
SLIPS I label this AN subs
PQON = PNQNA DeM
(DeMorgan's laws) in SLIPS
I label this ON subs
PQCN = PQNA
CN subs
Let's look at how we could change the second
problem to reflect these equivalences. Now there will be extra steps.
PQARSOC, RNSNA |- PNQNO
1. PQARSOC Ass
2. RNSNA
Ass
3. RSONPQANC 1
Contra
4. RSON
2 DeM
5. PQAN
3,4 C elim
6. PNQNO
5 DeM
Since we can work with equivalence rules even
inside another string, we could also have come up with this derivation.
1. PQARSOC Ass
2. RNSNA
Ass
3. RSONPQANC
1 Contra
4. RNSNAPQANC
3 DeM
5. RNSNAPNQNOC 4 DeM
6. PNQNO
2,5 C elim
Now we'll add a few additional inference rules to our list. Here, though, they apply only to the final letter in the string. We cannot use them inside a string, the way we do with equivalence rules.
P,Q |- PQA A
int
PQA |- P
A elim
P |- PQO
O int
PQO, PN |- Q
O elim
Now, let's use the basic pattern we've already
seen and make it still more complicated.
PQARSOC, QRNA, SN |- PN
1. PQARSOC Ass
2. QRNA
Ass
3. SN
Ass
4. RSONPQANC
1 Contra
5. RN
2 A elim
6. RNSNA
3,5 A int
7. RSON
6 DeM
8. PQAN
4,7 C elim
9. PNQNO
8 DeM
10. Q
2 A elim
11. QNN
10 DN
12. PN
9,11 O elim
Developing proficiency with derivations takes practice. This is why even this early in the course I am going to ask you to attempt some simple derivations using the set of rules we've presented so far. One way to approach a problem is to think of what you need to either introduce or eliminate, and then look at how by working with your premises you might get what you need.
Study the models above carefully before you go on to the assignment. In the assignment there will be different patterns, but you are to work step by step to go from what is given (Ass) to what you need to prove (the final line in the derivation).