demostrar que ( ∀ x {\displaystyle \forall _{x}} B ) ∧ {\displaystyle \wedge } A ≡ {\displaystyle \equiv } ∀ x {\displaystyle \forall _{x}} ( B ∧ {\displaystyle \wedge } A )
( ∀ x {\displaystyle \forall _{x}} B ) ∧ {\displaystyle \wedge } A ≡ {\displaystyle \equiv } ∀ x {\displaystyle \forall _{x}} B ∧ {\displaystyle \wedge } ∀ x {\displaystyle \forall _{x}} A -si x no esta libre en A
≡ {\displaystyle \equiv } ∀ x {\displaystyle \forall _{x}} ( B ∧ {\displaystyle \wedge } A )
NOTA: REVISAR TABLA (2.4) DE EQUIVALENCIAS QUE IMPLICAN CUANTIFICADORES SECCION 2.4 PAG 88
