Digging Deeper
Why do the rules of replacement work?
The two statements in a rule of replacement are logically equivalent, so given the same inputs, the resulting truth values will be the same
p ≡ ~~p
Double Negation (DN)
Use a rule of replacement on the entirety of:
(A ⊃ B) ⋅ (B ⊃ A)
A ≡ B BE
(B ⊃ A) ⋅ (A ⊃ B) Comm
1. (A v B) ⊃ C
2. B ∴ C
3. B v A
4. A v B
5. C
Add 2
Comm 3
MP 1, 4
1. A ⊃ (B ⋅ C)
2. (A ⊃ B) ⋅ (A ⊃ C) Dist 1
No! Distribution cannot be used with horseshoe
Why can rules of replacement be applied to part of a line?
Since the truth values come out the same for any given input, the truth value of the whole line will also be the same.
(p ⊃ q) ≡ (~q ⊃ ~p)
Contraposition (Contra)
Use a rule of replacement on the entirety of:
A
~~A DN
A v A Red
A ⋅ A Red
1. A ⊃ (B ⊃ C)
2. A
3. B ∴ C
4. A ⋅ B
5. (A ⋅ B) ⊃ C
6. C
Conj 2,3
Exp 1
MP 5,4
1. ~A ⊃ B
2. ~~A ⊃ B DN 1
No! Double negation always adds two curls - if it already has one, it will now have three, e.g. ~~~A ⊃ B
Give a line of reasoning that helps you remember a rule of replacement
Ex. Commutation, Association, and Double Negation work like math
Biconditional Exchange and the definition of ≡, left and right side of triple bar must both be true or both be false
~(p v q) ≡ (~p ⋅ ~q)
DeMorgan's (DeM)
Use a rule of replacement on the entirety of:
(A ⋅ B) v (A ⋅ ~B)
A ⋅ (B v ~B) Dist
1. A ≡ B
2. B ∴ A
3. (A ⊃ B) ⋅ (B ⊃ A)
4. (B ⊃ A) ⋅ (A ⊃ B)
5. B ⊃ A
6. A
BE 1
Comm 3
Simp 4
MP 5, 2
1. (A ⋅ B) ⋅ (A ⋅ C)
2. A ⋅ (B ⋅ C) Dist 1
No! But you can use Assoc, Comm, and Red to achieve the same result
Conditional Exchange says that (p ⊃ q) ≡ (~p v q). If I have:
1. A ⊃ B
2. A
What are two ways that I could get B? (You'll need rules of inference)
MP 1,2 and
3. ~A v B CE 1
4. B DS 3,2
(p ≡ q) ≡ [(p ⋅ q) v (~p ⋅ ~q)]
Biconditional Exchange (BE)
Use a rule of replacement on the entirety of:
(A ⊃ B) ⊃ (B ⊃ A)
~(A ⊃ B) v (B ⊃ A) CE
~(B ⊃ A) ⊃ ~(A ⊃ B) Contra
[(A ⊃ B) ⋅ B] ⊃ A Exp
1. A ≡ B
2. ~(A ⋅ B) ∴ ~(A v B)
3. (A ⋅ B) v (~A ⋅ ~B)
4. ~A ⋅ ~B
5. ~(A v B)
BE 1
DS 3,2
DeM 4
1. (A ⊃ B) ⊃ C
2. (A ⋅ B) ⊃ C Exp 1
No! Pay attention to the parentheses
Give an application of a rule of replacement that can also be achieved with a rule of inference
Ex: p to p v p
p ⋅ p to p
[(p ⋅ q) ⊃ r] ≡ [p ⊃ (q ⊃ r)]
Exportation (Exp)
Use a rule of replacement on the entirety of:
~{ [(A ⋅ B) ⋅ C] ⋅ [D ⋅ (E ⋅ F)] }
~[(A ⋅ B) ⋅ C] v ~[D ⋅ (E ⋅ F)] DeM
1. A ∴ (~A ⋅ B) ⊃ (C v D)
2. A v ~B
3. ~~A v ~B
4. ~(~A ⋅ B)
5. ~(~A ⋅ B) v (C v D)
6. (~A ⋅ B) ⊃ (C v D)
Add 1
DN 2
DeM 3
Add 4
CE 5
1. (A v B) v C
2. (C v B) v A Comm 1
No! You need intermediate steps because of the parentheses:
3. A v (B v C) Assoc 1
4. (B v C) v A Comm 3
5. (C v B) v A Comm 4