Theorem albi 785

Description: Split biconditional and distribute quantifier.

Assertion

Ref	Expression
albi	⊢ (∀x(φ ↔ ψ) ↔ (∀x(φ → ψ) ∧ ∀x(ψ → φ)))

Proof of Theorem albi

Step	Hyp	Ref	Expression
1		bi 396	. . 3 ⊢ ((φ ↔ ψ) ↔ ((φ → ψ) ∧ (ψ → φ)))
2	1	bial 695	. 2 ⊢ (∀x(φ ↔ ψ) ↔ ∀x((φ → ψ) ∧ (ψ → φ)))
3		19.26 749	. 2 ⊢ (∀x((φ → ψ) ∧ (ψ → φ)) ↔ (∀x(φ → ψ) ∧ ∀x(ψ → φ)))
4	2, 3	bitr 151	1 ⊢ (∀x(φ ↔ ψ) ↔ (∀x(φ → ψ) ∧ ∀x(ψ → φ)))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672

This theorem is referenced by:  hbbid 789  eu1 1019  eqss 1516  ssext 1865  cleqrel 2483  dmcosseq 2572

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677

This theorem depends on definitions:  df-bi 128  df-an 198

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