Theorem bibiad 32486

Description: Eliminate an hypothesis 𝜃 in a biconditional. (Contributed by Thierry Arnoux, 4-May-2025.)

Hypotheses

Ref	Expression
bibiad.1	⊢ ((𝜑 ∧ 𝜓) → 𝜃)
bibiad.2	⊢ ((𝜑 ∧ 𝜒) → 𝜃)
bibiad.3	⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
bibiad	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Proof of Theorem bibiad

Step	Hyp	Ref	Expression
1		simpl 482	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
2		bibiad.1	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜃)
3		simpr 484	. . 3 ⊢ ((𝜑 ∧ 𝜓) → 𝜓)
4		bibiad.3	. . . 4 ⊢ ((𝜑 ∧ 𝜃) → (𝜓 ↔ 𝜒))
5	4	biimpa 476	. . 3 ⊢ (((𝜑 ∧ 𝜃) ∧ 𝜓) → 𝜒)
6	1, 2, 3, 5	syl21anc 837	. 2 ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
7		simpl 482	. . 3 ⊢ ((𝜑 ∧ 𝜒) → 𝜑)
8		bibiad.2	. . 3 ⊢ ((𝜑 ∧ 𝜒) → 𝜃)
9		simpr 484	. . 3 ⊢ ((𝜑 ∧ 𝜒) → 𝜒)
10	4	biimpar 477	. . 3 ⊢ (((𝜑 ∧ 𝜃) ∧ 𝜒) → 𝜓)
11	7, 8, 9, 10	syl21anc 837	. 2 ⊢ ((𝜑 ∧ 𝜒) → 𝜓)
12	6, 11	impbida 800	1 ⊢ (𝜑 → (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207  df-an 396

This theorem is referenced by:  brab2d  32629  ellpi  33366