Theorem biadanid 614

Description: Deduction associated with biadani 612. Add a conjunction to an equivalence. (Contributed by Thierry Arnoux, 16-Jun-2024.)

Hypotheses

Ref	Expression
biadanid.1	⊢ ((𝜑 ∧ 𝜓) → 𝜒)
biadanid.2	⊢ ((𝜑 ∧ 𝜒) → (𝜓 ↔ 𝜃))

Assertion

Ref	Expression
biadanid	⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))

Proof of Theorem biadanid

Step	Hyp	Ref	Expression
1		biadanid.1	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
2		biadanid.2	. . . . . 6 ⊢ ((𝜑 ∧ 𝜒) → (𝜓 ↔ 𝜃))
3	2	biimpa 296	. . . . 5 ⊢ (((𝜑 ∧ 𝜒) ∧ 𝜓) → 𝜃)
4	3	an32s 568	. . . 4 ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) → 𝜃)
5	1, 4	mpdan 421	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜃)
6	1, 5	jca 306	. 2 ⊢ ((𝜑 ∧ 𝜓) → (𝜒 ∧ 𝜃))
7	2	biimpar 297	. . 3 ⊢ (((𝜑 ∧ 𝜒) ∧ 𝜃) → 𝜓)
8	7	anasss 399	. 2 ⊢ ((𝜑 ∧ (𝜒 ∧ 𝜃)) → 𝜓)
9	6, 8	impbida 596	1 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  dflidl2  14044  df2idl2  14065