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Question

 { D→(A∨¬C), ¬B→(D∨C), ¬A→¬(B∧C), (B∨D)↔C } ⊢A

Answer 1

Here's a derivation using the standard rules of inference and replacement along with conditional proof (denoted by vertical lines next to the step numbers). This is not the absolutely most efficient derivation (I used material equivalence on line 4 and then simplified in two different sub-proofs when I could have done it once outside the subproofs), but the best one is only going to be a bit shorter.

If you are using Fitch or SD then I would have to provide a completely different derivation using those rules.

If you're interested in doing some lessons, let me know.

1. D→(A∨¬C)

2. ¬B→(D∨C)

3. ¬A→¬(B∧C)

4. (B∨D)↔C .............. /A

5. | B..........................ACP

6 | B v D.....................Add 9

7 | [(B∨D) →C] ∧ [C →(B∨D)] ME 4

8 | (B∨D) →C..............Simp 7

9 | C.............................MP 6, 8

10| B∧C.......................Conj. 5, 9

11| ¬¬(B∧C).................DN 10

12| ¬¬A........................MT 3, 11

13| A ...........................DN 12

14 B→A......................CP 5-13

15| ¬B...........................ACP

16| (D v C).....................MP 2, 15

17|| D............................ACP

18|| D ∨ B....................,Add 17

19|| B∨D.......................Comm 18

20|| [(B∨D) →C] ∧ [C →(B∨D)] ME 4

21|| [(B∨D) →C..............Simp 20

22|| C..............................MP 19, 21

23|| ¬¬C..........................DN 22

24|| A v ¬C......................MP 1, 17

25|| A...............................DS 23, 24

26| D → A........................CP 17-25

27|| C...............................ACP

28|| (B∨D) →C] ∧ [C →(B∨D)] ME 4

29|| C →(B∨D).................Simp 28

30|| B v D.........................MP 27, 29

31|| D...............................DS 15, 30

32|| A................................MP 26, 31

33| C→A............................CP 27-32

34| (D→A) ∧ (C → A)........Conj 26, 33

35| (D v C) →(A v A)..........CD 34

36| A v A...........................MP16, 35

37| A.................................Taut 36

38 ¬B →A..........................CP 15-37

39 (¬B →A) ∧ (B →A)........Conj 14, 38

40 (¬B v B) →(A v A)..........CD 39

41 | B..................................ACP

42 | B v B............................Taut 41

43 | B..................................Taut 42

44 B→B..............................CP 41-43

45 ¬B v B...........................MI 44

46 A v A.............................MP40, 45

47 A...................................Taut 46 QED

Answer 2

If B holds true, then B or D also holds true....

Then by the third given statement, C must also hold hold true...

So B and C are both true.

The contrapositive of the second given statement then says A must hold true

(end of proof)