Theorem 2albiim

Description: Split a biconditional and distribute two quantifiers. (Contributed by NM, 3-Feb-2005)

	Ref	Expression
Assertion	2albiim	$$⊢\forall x\forall y\left(\phi \leftrightarrow \psi \right)\leftrightarrow \left(\forall x\forall y\left(\phi \to \psi \right)\wedge \forall x\forall y\left(\psi \to \phi \right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		albiim	$$⊢\forall y\left(\phi \leftrightarrow \psi \right)\leftrightarrow \left(\forall y\left(\phi \to \psi \right)\wedge \forall y\left(\psi \to \phi \right)\right)$$
2	1	albii	$$⊢\forall x\forall y\left(\phi \leftrightarrow \psi \right)\leftrightarrow \forall x\left(\forall y\left(\phi \to \psi \right)\wedge \forall y\left(\psi \to \phi \right)\right)$$
3		19.26	$$⊢\forall x\left(\forall y\left(\phi \to \psi \right)\wedge \forall y\left(\psi \to \phi \right)\right)\leftrightarrow \left(\forall x\forall y\left(\phi \to \psi \right)\wedge \forall x\forall y\left(\psi \to \phi \right)\right)$$
4	2 3	bitri	$$⊢\forall x\forall y\left(\phi \leftrightarrow \psi \right)\leftrightarrow \left(\forall x\forall y\left(\phi \to \psi \right)\wedge \forall x\forall y\left(\psi \to \phi \right)\right)$$