Final Answer:
Using both Conditional Proof and Indirect Proof, we have derived ∼H ∨ ∼L from premises 1 and 2.
Step-by-step explanation:
To prove ∼H ∨ ∼L from the premises using Conditional Proof (CP) and Indirect Proof (IP), let's break it down step by step:
- H ⊃ (L ⊃ K) (Premise)
- L ⊃ (K ⊃ ∼L) (Premise)
We want to derive ∼H ∨ ∼L using CP or IP.
Using Conditional Proof (CP):
Assume H (Assumption for CP)
- L ⊃ K (From 1, 3; Modus Ponens)
- K ⊃ ∼L (From 2, Modus Ponens)
- K (From 4, 3; Modus Ponens)
- ∼L (From 6, 5; Modus Ponens)
- ∼H ∨ ∼L (From 3-7; Disjunction Introduction)
Now, let's use Indirect Proof (IP) to derive ∼H ∨ ∼L:
Assume ∼(∼H ∨ ∼L) (Assumption for IP)
- ∼H ∧ L (Assume ∼(∼H ∨ ∼L) and use De Morgan's Law)
- L (From 10; Simplification)
- K (From 2, 11; Modus Ponens)
- H (Assume ∼(∼H ∨ ∼L) and use Reductio ad Absurdum)
- ∼H (From 9-13; Indirect Proof)
- ∼H ∨ ∼L (From 14; Addition)
Therefore, using both Conditional Proof and Indirect Proof, we have derived ∼H ∨ ∼L from premises 1 and 2.