Theorem 3anbi123d 1252

Description: Deduction joining 3 equivalences to form equivalence of conjunctions. (Contributed by NM, 22-Apr-1994.)

Hypotheses

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

Assertion

Ref	Expression
3anbi123d	⊢ (φ → ((ψ ∧ θ ∧ η) ↔ (χ ∧ τ ∧ ζ)))

Proof of Theorem 3anbi123d

Step	Hyp	Ref	Expression
1		bi3d.1	. . . 4 ⊢ (φ → (ψ ↔ χ))
2		bi3d.2	. . . 4 ⊢ (φ → (θ ↔ τ))
3	1, 2	anbi12d 691	. . 3 ⊢ (φ → ((ψ ∧ θ) ↔ (χ ∧ τ)))
4		bi3d.3	. . 3 ⊢ (φ → (η ↔ ζ))
5	3, 4	anbi12d 691	. 2 ⊢ (φ → (((ψ ∧ θ) ∧ η) ↔ ((χ ∧ τ) ∧ ζ)))
6		df-3an 936	. 2 ⊢ ((ψ ∧ θ ∧ η) ↔ ((ψ ∧ θ) ∧ η))
7		df-3an 936	. 2 ⊢ ((χ ∧ τ ∧ ζ) ↔ ((χ ∧ τ) ∧ ζ))
8	5, 6, 7	3bitr4g 279	1 ⊢ (φ → ((ψ ∧ θ ∧ η) ↔ (χ ∧ τ ∧ ζ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   ∧ w3a 934

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360  df-3an 936

This theorem is used by:  3anbi12d  1253  3anbi13d  1254  3anbi23d  1255  ax11wdemo  1723  sbc3ang  3105  pw1equn  4332  pw1eqadj  4333  cenc  6182