Theorem alsbid 50867

Description: Deduction form of alsbii 50865. (Contributed by David A. Wheeler, 12-Jul-2026.)

Hypotheses

Ref	Expression
alsbid.1	⊢ Ⅎ𝑥𝜑
alsbid.2	⊢ (𝜑 → (𝜓 ↔ 𝜃))
alsbid.3	⊢ (𝜑 → (𝜒 ↔ 𝜏))

Assertion

Ref	Expression
alsbid	⊢ (𝜑 → (∀∃𝑥(𝜓 → 𝜒) ↔ ∀∃𝑥(𝜃 → 𝜏)))

Proof of Theorem alsbid

Step	Hyp	Ref	Expression
1		alsbid.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2		alsbid.2	. . . . 5 ⊢ (𝜑 → (𝜓 ↔ 𝜃))
3		alsbid.3	. . . . 5 ⊢ (𝜑 → (𝜒 ↔ 𝜏))
4	2, 3	imbi12d 347	. . . 4 ⊢ (𝜑 → ((𝜓 → 𝜒) ↔ (𝜃 → 𝜏)))
5	1, 4	albid 2259	. . 3 ⊢ (𝜑 → (∀𝑥(𝜓 → 𝜒) ↔ ∀𝑥(𝜃 → 𝜏)))
6	1, 2	exbid 2260	. . 3 ⊢ (𝜑 → (∃𝑥𝜓 ↔ ∃𝑥𝜃))
7	5, 6	anbi12d 644	. 2 ⊢ (𝜑 → ((∀𝑥(𝜓 → 𝜒) ∧ ∃𝑥𝜓) ↔ (∀𝑥(𝜃 → 𝜏) ∧ ∃𝑥𝜃)))
8		df-als 50853	. 2 ⊢ (∀∃𝑥(𝜓 → 𝜒) ↔ (∀𝑥(𝜓 → 𝜒) ∧ ∃𝑥𝜓))
9		df-als 50853	. 2 ⊢ (∀∃𝑥(𝜃 → 𝜏) ↔ (∀𝑥(𝜃 → 𝜏) ∧ ∃𝑥𝜃))
10	7, 8, 9	3bitr4g 317	1 ⊢ (𝜑 → (∀∃𝑥(𝜓 → 𝜒) ↔ ∀∃𝑥(𝜃 → 𝜏)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812  Ⅎwnf 1816  ∀∃wals 50851

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2213

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-als 50853

This theorem is used by: (None)