Proving Symbolic Logic

Question

Prove that the following is valid:

b ∧ a

a → c

(c ∧ ¬d) → ¬b

∴ d

So far I have only done:

b (Rule of Conjunctive Simplification on b ∧ a)

b ∨ (c ∧ ¬d) (Rule of Disjunctive Amplification on b)

I'm not really sure what to do next please help me out.

Answer 1

$1.\quad b\land a$

$2.\quad a\to c$

$3.\quad(c\land\lnot d)\to\lnot b$

$4.\quad b\quad$ (Simplify 1)

$5.\quad a\quad$ (Simplify 1)

$6.\quad c\quad$ (Modus ponens 5,2)

$7.\quad\lnot\lnot b\to\lnot(c\land\lnot d)\quad$ (Contraposition 3)

$8.\quad b\to\lnot(c\land\lnot d)\quad$ (Double negation 7)

$9.\quad\lnot(c\land\lnot d)\quad$ (Modus ponens 4,8)

$10.\quad\lnot c\lor\lnot\lnot d\quad$ (DeMorgan 9)

$11.\quad c\to\lnot\lnot d\quad$ (Implication 10)

$12.\quad\lnot\lnot d\quad$ (Modus ponens 6,11)

$13.\quad d\quad$ (Double negation 12)

Answer 2

You wish to prove $d$ is entailed by the premises.   Well, hopefully you may derive a contradiction by accepting the premises and assuming $\lnot d$.   That is to say: you should try for a proof by contradiction.

Clearly the third premise will be useful for this, if we can first derive $c$ and $b$.

Now $b$ may be derived from the first premise, as might $a$.   Well, $c$ can then be derived from that $a$ and the second premise.

$$\def\fitch#1#2{}\fitch{~1.~b\land a\qquad\textsf{P1}\\~2.~a\to c\qquad\textsf{P2}\\~3.~(c\land\lnot d)\to\lnot b\qquad\textsf{P3}}{~4.~b\quad\land\textsf{E},1\\~5.~a\quad\land\textsf{E},1\\~6.~c\quad\to\textsf{E},4,2\\\fitch{~7.~\lnot d\quad\textsf{A}}{~8.~c\land\lnot d\quad\land\textsf{I},6,7\\~9.~\lnot b\quad\to\textsf{E},8,3\\10.~\bot\quad\lnot\textsf{E},4,9}\\11.~\lnot\lnot d\quad\lnot\textsf{I},7{-}10\\12.~d\quad\lnot\lnot\textsf{E},11}$$