propositional logic?

Question

I need to proove this arument valid using the rules of inference. I'm stuck, can someone help me? I don't have a keyboard with logic symbols, so I hope this makes sense.

1. ~(A dot B) triple bar ~C
2. (D v E) horeshoe C / E horseshoe A

Answer 1

Dot/triple bar/horseshoe... sounds like Hurley notation.

Below is a Hurley-style direct proof (conditional or indirect proof would be different than this direct proof).

& = dot
<-> = triple bar
-> = horseshoe

1. ~(A & B) <-> ~C
2. (D v E) -> C..................../E -> A
3. [~(A & B) -> ~C] & [~C -> ~(A & B)]...1, Equiv
4. ~(A & B) -> ~C...............3, Simp
5. ~~C -> ~~(A & B)...........4, Trans
6. C -> ~~(A & B)...............5, DN
7. C -> (A & B)...................6, DN
8. ~C v (A & B)...................7, Impl
9. (~C v A) & (~C v B)........8, Dist
10. ~C v A........................9, Simp
11. C -> A.......................10, Impl
12. ~(D v E) v C................2, Impl
13. (~D & ~E) v C.......... 12, DM
14. C v (~D & ~E)...........13, Com
15. (C v ~D) & (C v ~E)....14, Dist
16. (C v ~E) & (C v ~D)....15, Com
17. C v ~E......................16, Simp
18. ~E v C......................17, Com
19. E -> C.......................18, Impl
20. E -> A.......................11,19, HS

Rules:
Equiv: Material Equivalence
Simp: Simplification
Trans: Transposition
DN: Double Negation
Impl: Material Implication
Dist: Distribution
Simp: Simplification
DM: DeMorgans' Rule
Com: Commutativity
HS: Hypothetical Syllogism

Answer 2

I hope you'll forgive me if I give the answer in more modern notation; ⊃ is too easily confused with the symbol for proper superset. I'll assume you're using a natural deduction system like the one in the linked Wikipedia article, although note that there are significant variations in the particular form of the axioms, so check your textbook to see whether this matches the particular format you are supposed to use:

1: E -- hypothesis
1.1: D∨E -- from 1 by disjunction introduction
1.2: (D∨E) → C -- premise 2
1.3: C -- from 1.1 and 1.2 by modus ponens
1.4: ¬(A∧B) -- hypothesis
1.4.1: C -- reiteration of 1.3
1.5: ¬(A∧B) → C -- from 1.4 through 1.4.1 by conditional proof
1.6: ¬(A∧B) ↔ ¬C -- premise 1
1.7: ¬(A∧B) → ¬C -- from 1.6 by biconditional elimination
1.8: ¬¬(A∧B) -- from 1.5 and 1.7 by reductio ad absurdum
1.9: A∧B -- from 1.8 by double negation elimination
1.10: A -- from 1.9 by conjunction elimination
2: E → A -- from 1 through 1.10 by conditional proof