Here are some tricks:
(1) Whenever you see that you have inconsistent premises, you can derive any wff whatever.
P, ~P |- Q
1. P
Ass
2. ~P
Ass
3. P v Q 1 v int
4. Q
2,3 v elim
Think why. All argument forms are either
valid or invalid, but by definition a form cannot be invalid unless all
premises are true.
(2) Whenever you are trying to establish a conditional and you have either a false antecedent or a true consequent in the premises, you can make use of our "magic rule" to introduce a disjunction.
~P |- P -> Q
1. ~P
Ass
2. ~P v Q 1
v int
3. P -> Q 2
MI
Q |- P->Q
1. Q
Ass
2. ~P v Q 1 v int
3. P -> Q 2 MI
(3) You can always convert a conjunction into a conditional (but not the other way around).
P & Q |- P -> Q
1. P & Q
Ass
2. Q
1 & elim
3. ~P v Q 2
v int
4. P -> Q 3
MI
And here are some traps.
(1) "And" sometimes should be symbolized as a disjunction.
Symbolize "All cats and dogs are to be
kept in a kennel."
Ax((Cx v Dx) -> Kx)
Using a conjunction would say we has strange
creature that was both a cat and a dog.
(2) The new name rule warns us to work with existential quantifiers before we work with universal quantifiers whenever we can. This is important in working with consistency trees, since often we negate a universal quantifier, which means we now have an existential one.
Ax(Fx)->Ax(Gx) |- Ax(Fx -> Gx)
(3) We must always remember to work with the main connective in applying an inference rule, and this is especially true whenever we are introducing or eliminating quantifiers,
This is a proof that does not work, even though the argument form is valid:
Ex(Fx) v Ax(Gx) |- Ex(Fx v Gx)
1. Ex(Fx) v Ax(Gx)
Ass
2. Fa v Ax(Gx)
1 EQ elim error
3. Fa v Ga
2 UQ elim error
4. Ex(Fx v Gx)
3 EQ int
For an acceptable proof we would need to do an indirect derivation:
1. Ex(Fx) v Ax(Gx)
Ass
2.
~Ex(Fx v Gx) Hyp
3.
Ax ~(Fx v Gx) 2 QN
4.
Ax(~Fx & ~Gx) 3 DeM
5.
~Fa & ~Ga 4
UQ elim
6.
~Fa
5 & elim
7.
Ax(~Fx)
6 UQ int note why "EQ int" would not work
8.
~Ex(Fx)
7 QN
9.
Ax(Gx)
1,8 v elim
10. Ga
9 UQ elim
11. ~Ga
5 & elim
12. Ga &
~Ga
10,11 & int
13. Ex(Fx v Gx)
2-12 hyp elim
Obviously a longer proof, but one that is correctly done.