I'm trying to prove this argument but I can't seem to find a way to prove it.
$a \to b$
$b \lor c$
$(c \land \sim a) \to (d \land \sim a)$
$\sim b$
$\therefore d$
EDIT: Attached below is the rules we are allowed to use. Rules
I'm trying to prove this argument but I can't seem to find a way to prove it.
$a \to b$
$b \lor c$
$(c \land \sim a) \to (d \land \sim a)$
$\sim b$
$\therefore d$
EDIT: Attached below is the rules we are allowed to use. Rules
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$1)$ $\overline{b}\to c$ (Given Premise).
$2)$ $\overline{b}$ (premise)
$3)$ $c$ (Modus ponens of 1,2)
$4)$ $\overline{b} \to \overline{a}$ (premise)
$5)$ $\overline{a}$ (Modus ponens of 2 ,4)
$6)$ $c$ $\wedge$ $\overline{a}$ (Conjunction of 3 ,5)
$7)$ $(c$ $\wedge$ $\overline{a})$ $\implies$ $(d$ $\wedge$ $\overline{a})$ (premise)
$8)$ $d$ $\wedge$ $\overline{a}$ (Modus ponens of 6,7)
$9)$ $d$ (From 8).