I need help with this proof for my philosophy class. This proof is supposed to be done via indirect proof or conditional proof, so it is supposed to use AIP and IP or ACP and CP to derive the conclusion! This is an assignment that is submitted through Aplia, so I need it to be precise and the assignment is due tonight, so I would like it done before tonight please!
This is the key:
The * = dot
The > = horseshoe
The v = wedge
The ≡ = triple bar
The ~ = tilde
The statement after the / (slash) is the conclusion that I need.
The problem is:
1. R ≡ ~O
/ (~T > M) v (T > ~H)
To solve this proof using indirect proof or conditional proof, we'll walk through the steps together.
1. Start by assuming the opposite of the conclusion as the assumption to work with.
Assume: ~(~T > M) v (T > ~H)
2. Apply DeMorgan's law to the assumption to simplify it.
~(~T > M) v (T > ~H) => (T ∧ ~M) v (~T ∧ H)
3. Start a subproof to derive a contradiction, based on the assumption in step 1.
4. Assume the opposite of what you want to prove.
Assume: ~(~(T > M) v (T > ~H))
5. Apply DeMorgan's law to simplify the assumption.
~(~(T > M) v (T > ~H)) => ~~(T > M) ∧ ~~(T > ~H)
=> (T > M) ∧ (T > ~H)
6. Use the biconditional equivalence (≡) in premise 1 to break down (R ≡ ~O).
R ≡ ~O => (~O > R) ∧ (R > ~O)
7. Separate the two parts of the biconditional in step 6.
(~O > R) ∧ (R > ~O) => (~O > R) and (R > ~O)
8. Use the transposition equivalence to convert implication (P > Q) into its contrapositive (~Q > ~P).
(~O > R) => (~~R > ~~O) => (R > O)
9. Apply simplification to the conjunction (T > O) ∧ (R > O).
(T > O) ∧ (R > O) => T > O
10. Derive a contradiction by applying Modus Ponens to steps 8 and 9.
(R > O) ∧ (T > O)
O # Contradiction
11. Close the subproof and negate the assumption to complete the indirect proof.
~(~(T > M) v (T > ~H))
~~(T > M) ∧ ~~(T > ~H)
(T > M) ∧ (T > ~H)
T > M (Simplification)
M (Modus Ponens)
~M # Contradiction
4. Close the subproof and finish the proof by using disjunction introduction to get the final conclusion.
~(~T > M) v (T > ~H)
(~T > M) > (~ ~T > M) # Conditional Proof (CP)
(~T > M) > (T > M) # Double Negation
~M v (T > M) # Disjunction Introduction (DI)
(T > M) v ~M # Commutation
(~T > M) v ~M # Commutation
(~T > M) v (~T > ~H) # Commutation
(~T > M) v (T > ~H) # Distribution (DI)
Thus, we have derived the conclusion (~T > M) v (T > ~H) using indirect proof or conditional proof.