Using Rules of Inference, can you show step by step that this argument is valid? NOT(IF p THEN q) AND p = NOT(q)

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1)NOT(IF p THEN q) Premise

2)p Premise

3)p AND NOT(q) Negating a Conditional(1)

4)NOT(q) Conjunction elimination(3)

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Solution: a) p∧q b)p∧¬q c)¬p∧¬ q d)p∨q e)p→ q f)(p∨q)∧(p→ ¬ q) g)p↔q

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