Using inference rules/equivalences

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The discussion focuses on using inference rules and equivalences to demonstrate that the expression ((¬ A ∨ ¬ B) → (C ∧ D)) ∧ (C → E) ∧ (¬ E) leads to A. The initial steps involve simplifying the expression, noting that the conjunction of ¬E with the implications provides necessary conditions for A. A key point is recognizing that the simplification leads to the conclusion that both ¬C and ¬E must hold, which implies that A must be true. The final argument hinges on the logical structure of the implications and their negations. This analysis emphasizes the importance of correctly applying inference rules to derive conclusions in logical expressions.
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Homework Statement


Using[/PLAIN] inference rules/equivalencies, Show that
( ( (¬ A ∨ ¬ B ) → ( C ∧ D ) ) ∧ ( C → E) ∧ ( ¬ E )) → A


Homework Equations


( ( (¬ A ∨ ¬ B ) → ( C ∧ D ) ) ∧ ( C → E) ∧ ( ¬ E )) → A[/B]

The Attempt at a Solution



Using inference rules/equivalencies, Show that
( ( (¬ A ∨ ¬ B ) → ( C ∧ D ) ) ∧ ( C → E) ∧ ( ¬ E )) → A

This is my answer.
Consider about,
( (¬ A ∨ ¬ B ) → ( C ∧ D ) ) ∧ ( C → E) ∧ ( ¬ E )
Then,
( (¬ A ∨ ¬ B ) → ( C ∧ D ) ) ∧ ¬ E
( ¬( A ∧ B ) → ( C ∧ D ) ) ∧ ¬ E
( ¬¬( A ∧ B ) ∨ ( C ∧ D ) ) ∧ ¬ E
( ( A ∧ B ) ∨ ( C ∧ D ) ) ∧ ¬ E
After that, I cannot reach the answer → A
 
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To make sure I am reading this right, I will use English connectors.
[ (not A or nor B ) implies (C and D) ] and [( C implies E ) and (not E) ].
The right side simplifies to (not C) and (not E), not just (not E) like you have.
Having the not C will help you conclude that A must be true.
 
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