Theorem bi3and 636

Description: Deduction joining 3 equivalences to form equivalence of conjunctions.

Hypotheses

Ref	Expression
bi3d.1	⊢ (φ → (ψ ↔ χ))
bi3d.2	⊢ (φ → (θ ↔ τ))
bi3d.3	⊢ (φ → (η ↔ ζ))

Assertion

Ref	Expression
bi3and	⊢ (φ → ((ψ ∧ θ ∧ η) ↔ (χ ∧ τ ∧ ζ)))

Proof of Theorem bi3and

Step	Hyp	Ref	Expression
1		bi3d.1	. . . 4 ⊢ (φ → (ψ ↔ χ))
2		bi3d.2	. . . 4 ⊢ (φ → (θ ↔ τ))
3	1, 2	anbi12d 476	. . 3 ⊢ (φ → ((ψ ∧ θ) ↔ (χ ∧ τ)))
4		bi3d.3	. . 3 ⊢ (φ → (η ↔ ζ))
5	3, 4	anbi12d 476	. 2 ⊢ (φ → (((ψ ∧ θ) ∧ η) ↔ ((χ ∧ τ) ∧ ζ)))
6		df-3an 583	. 2 ⊢ ((ψ ∧ θ ∧ η) ↔ ((ψ ∧ θ) ∧ η))
7		df-3an 583	. 2 ⊢ ((χ ∧ τ ∧ ζ) ↔ ((χ ∧ τ) ∧ ζ))
8	5, 6, 7	3bitr4g 428	1 ⊢ (φ → ((ψ ∧ θ ∧ η) ↔ (χ ∧ τ ∧ ζ)))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196   ∧ w3a 581

This theorem is referenced by:  so 2152  limeq 2211  tz9.1 3490  mulcant2 4209  sqrlem20 4750

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

This theorem depends on definitions:  df-bi 128  df-an 198  df-3an 583

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