Theorem 3anbi123i 1140

Description: Join 3 biconditionals with conjunction. (Contributed by NM, 21-Apr-1994.)

Hypotheses

Ref	Expression
bi3.1	⊢ (φ ↔ ψ)
bi3.2	⊢ (χ ↔ θ)
bi3.3	⊢ (τ ↔ η)

Assertion

Ref	Expression
3anbi123i	⊢ ((φ ∧ χ ∧ τ) ↔ (ψ ∧ θ ∧ η))

Proof of Theorem 3anbi123i

Step	Hyp	Ref	Expression
1		bi3.1	. . . 4 ⊢ (φ ↔ ψ)
2		bi3.2	. . . 4 ⊢ (χ ↔ θ)
3	1, 2	anbi12i 678	. . 3 ⊢ ((φ ∧ χ) ↔ (ψ ∧ θ))
4		bi3.3	. . 3 ⊢ (τ ↔ η)
5	3, 4	anbi12i 678	. 2 ⊢ (((φ ∧ χ) ∧ τ) ↔ ((ψ ∧ θ) ∧ η))
6		df-3an 936	. 2 ⊢ ((φ ∧ χ ∧ τ) ↔ ((φ ∧ χ) ∧ τ))
7		df-3an 936	. 2 ⊢ ((ψ ∧ θ ∧ η) ↔ ((ψ ∧ θ) ∧ η))
8	5, 6, 7	3bitr4i 268	1 ⊢ ((φ ∧ χ ∧ τ) ↔ (ψ ∧ θ ∧ η))

This proof depends on syntax axioms:   ↔ 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:  3anbi1i  1142  3anbi2i  1143  3anbi3i  1144  syl3anb  1225  cadnot  1394  opksnelsik  4266  eloprabga  5579  restxp  5787  oqelins4  5795  xpassen  6058  mucass  6136  taddc  6230  letc  6232