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Theorem disjeq1i 36981
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.
StepHypRef Expression
1 disjeq1i.1 . . . 4 𝐴 = 𝐵
21rmoeqi 36976 . . 3 (∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
32albii 1851 . 2 (∀𝑡∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
4 df-disj 5070 . 2 (Disj 𝑥 ∈ 𝐴 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶)
5 df-disj 5070 . 2 (Disj 𝑥 ∈ 𝐵 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
63, 4, 53bitr4i 306 1 (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1567   = wceq 1569   ∈ wcel 2144  ∃*wrmo 3364  Disj wdisj 5069
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-mo 2564  df-cleq 2752  df-clel 2835  df-rmo 3365  df-disj 5070
This theorem is used by:  disjeq12i  36982
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