Theorem alsbii 51079

Description: Congruence: equivalents may be substituted inside an "all some". (Contributed by David A. Wheeler, 12-Jul-2026.)

Hypotheses

Ref	Expression
alsbii.1	⊢ (𝜑 ↔ 𝜒)
alsbii.2	⊢ (𝜓 ↔ 𝜃)

Assertion

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

Proof of Theorem alsbii

Step	Hyp	Ref	Expression
1		alsbii.1	. . . . 5 ⊢ (𝜑 ↔ 𝜒)
2		alsbii.2	. . . . 5 ⊢ (𝜓 ↔ 𝜃)
3	1, 2	imbi12i 353	. . . 4 ⊢ ((𝜑 → 𝜓) ↔ (𝜒 → 𝜃))
4	3	albii 1851	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) ↔ ∀𝑥(𝜒 → 𝜃))
5	1	exbii 1880	. . 3 ⊢ (∃𝑥𝜑 ↔ ∃𝑥𝜒)
6	4, 5	anbi12i 640	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑) ↔ (∀𝑥(𝜒 → 𝜃) ∧ ∃𝑥𝜒))
7		df-als 51067	. 2 ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∃𝑥𝜑))
8		df-als 51067	. 2 ⊢ (∀∃𝑥(𝜒 → 𝜃) ↔ (∀𝑥(𝜒 → 𝜃) ∧ ∃𝑥𝜒))
9	6, 7, 8	3bitr4i 306	1 ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ ∀∃𝑥(𝜒 → 𝜃))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811  ∀∃wals 51065

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-als 51067

This theorem is used by:  2alsraln0  51096