Theorem disjeq1i 36799

Description: Equality theorem for disjoint collection. Inference version. (Contributed by GG, 1-Sep-2025.)

Hypothesis

Ref	Expression
disjeq1i.1	⊢ 𝐴 = 𝐵

Assertion

Ref	Expression
disjeq1i	⊢ (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶)

Proof of Theorem disjeq1i

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		disjeq1i.1	. . . 4 ⊢ 𝐴 = 𝐵
2	1	rmoeqi 36794	. . 3 ⊢ (∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
3	2	albii 1852	. 2 ⊢ (∀𝑡∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
4		df-disj 5075	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶)
5		df-disj 5075	. 2 ⊢ (Disj 𝑥 ∈ 𝐵 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
6	3, 4, 5	3bitr4i 306	1 ⊢ (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶)

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570   ∈ wcel 2145  ∃*wrmo 3366  Disj wdisj 5074

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-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-mo 2566  df-cleq 2754  df-clel 2837  df-rmo 3367  df-disj 5075

This theorem is used by:  disjeq12i  36800