Theorem ralbiim 2752

Description: Split a biconditional and distribute quantifier. (Contributed by NM, 3-Jun-2012.)

Assertion

Ref	Expression
ralbiim	⊢ (∀x ∈ A (φ ↔ ψ) ↔ (∀x ∈ A (φ → ψ) ∧ ∀x ∈ A (ψ → φ)))

Proof of Theorem ralbiim

Step	Hyp	Ref	Expression
1		dfbi2 609	. . 3 ⊢ ((φ ↔ ψ) ↔ ((φ → ψ) ∧ (ψ → φ)))
2	1	ralbii 2639	. 2 ⊢ (∀x ∈ A (φ ↔ ψ) ↔ ∀x ∈ A ((φ → ψ) ∧ (ψ → φ)))
3		r19.26 2747	. 2 ⊢ (∀x ∈ A ((φ → ψ) ∧ (ψ → φ)) ↔ (∀x ∈ A (φ → ψ) ∧ ∀x ∈ A (ψ → φ)))
4	2, 3	bitri 240	1 ⊢ (∀x ∈ A (φ ↔ ψ) ↔ (∀x ∈ A (φ → ψ) ∧ ∀x ∈ A (ψ → φ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wral 2615

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-ral 2620

This theorem is used by:  eqreu  3029  ssofeq  4078