Formal Proving Discrete maths Prove using logical equivalence rules/laws that
$$~[ ∧ ~ ∧ ( ∨ )] ∨ (~ ∧ ) ∧ ~( ∨ ~ ∨ ~) ≡ ~ ∨ b$$
 A: I'll prove the left to right half using a semantic argument, have your truth tables ready to check my reasoning:

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*A basic principle of disjunction (POD):

$$\Gamma, \phi \lor \psi \vDash \chi \;\;\; \text{iff} \;\;\; \Gamma,\phi \vDash \chi \,\,\, \text{and} \,\,\, \Gamma,\psi \vDash \chi$$


*A basic principle of conjunction (POC):

$$\Gamma \vDash \phi \land \psi \;\;\; \text{iff} \;\;\; \Gamma \vDash \phi \,\,\, \text{and} \,\,\, \Gamma \vDash \psi$$


*Deduction Teorem (DT):

$$\Gamma \vDash \phi \to \psi \;\;\; \text{iff} \;\;\; \Gamma,\phi\vDash\psi$$
Also: POD implies this introduction rule:
$$\phi \vDash \phi \lor \psi$$
And obviously assumption (Ass):
$$ \phi \vDash \phi$$

First consider the Assumption: $\Gamma,[a \land b \land (a \lor c) ]\vDash [(a \land b) \land (a \lor c) ]$; by POC we get: $$\Gamma,[a \land b \land (a \lor c) ]\vDash a\land b$$ and with another application of POC we obtain: $$\Gamma,[(a \land b) \land (a \lor c) ]\vDash a$$
Therefore, using the introduction rule: $$\Gamma,[(a \land b) \land (a \lor c) ]\vDash a \lor b.$$
Now consider (Ass): $\Gamma,[( a \land b) \land (a \lor b \lor c)] \vDash( a \land b) \land (a \lor b \lor c)$; with a simple use of POC we obtain: $$\Gamma, [( a \land b) \land (a \lor b \lor c)]\vDash a\land b$$
And thus with an equivalent argument we deduce: $$\Gamma,[( a \land b) \land (a \lor b \lor c)]\vDash a \lor b.$$
Now we use POD to obtain the formula:
$$\Gamma, \, [a \land b \land (a \lor c) ] \, \lor \, [( a \land b) \land (a \lor b \lor c)] \vDash a \lor b  $$
And finally we use the deduction theorem to get the first result we want:
$$\Gamma \vDash \{[a \land b \land (a \lor c) ] \, \lor \, [( a \land b) \land (a \lor b \lor c)]\} \to a \lor b$$

Now you must start from $a \lor b$ and deduce $[a \land b \land (a \lor c) ] \, \lor \, [( a \land b) \land (a \lor b \lor c)]$.
