Theorem disjeq12dv 37004

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

Hypotheses

Ref	Expression
disjeq12dv.1	⊢ (𝜑 → 𝐴 = 𝐵)
disjeq12dv.2	⊢ (𝜑 → 𝐶 = 𝐷)

Assertion

Ref	Expression
disjeq12dv	⊢ (𝜑 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐷))

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)   𝐶(𝑥)   𝐷(𝑥)

Proof of Theorem disjeq12dv

Dummy variable 𝑡 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		disjeq12dv.1	. . . . . . . 8 ⊢ (𝜑 → 𝐴 = 𝐵)
2	1	eleq2d 2847	. . . . . . 7 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	2	anbi1d 643	. . . . . 6 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝑡 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑡 ∈ 𝐶)))
4	3	mobidv 2575	. . . . 5 ⊢ (𝜑 → (∃*𝑥(𝑥 ∈ 𝐴 ∧ 𝑡 ∈ 𝐶) ↔ ∃*𝑥(𝑥 ∈ 𝐵 ∧ 𝑡 ∈ 𝐶)))
5		df-rmo 3366	. . . . 5 ⊢ (∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∃*𝑥(𝑥 ∈ 𝐴 ∧ 𝑡 ∈ 𝐶))
6		df-rmo 3366	. . . . 5 ⊢ (∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶 ↔ ∃*𝑥(𝑥 ∈ 𝐵 ∧ 𝑡 ∈ 𝐶))
7	4, 5, 6	3bitr4g 317	. . . 4 ⊢ (𝜑 → (∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶))
8	7	albidv 1953	. . 3 ⊢ (𝜑 → (∀𝑡∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶))
9		df-disj 5071	. . 3 ⊢ (Disj 𝑥 ∈ 𝐴 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐴 𝑡 ∈ 𝐶)
10		df-disj 5071	. . 3 ⊢ (Disj 𝑥 ∈ 𝐵 𝐶 ↔ ∀𝑡∃*𝑥 ∈ 𝐵 𝑡 ∈ 𝐶)
11	8, 9, 10	3bitr4g 317	. 2 ⊢ (𝜑 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))
12		disjeq12dv.2	. . . 4 ⊢ (𝜑 → 𝐶 = 𝐷)
13	12	adantr 486	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐵) → 𝐶 = 𝐷)
14	13	disjeq2dv 5075	. 2 ⊢ (𝜑 → (Disj 𝑥 ∈ 𝐵 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐷))
15	11, 14	bitrd 282	1 ⊢ (𝜑 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐷))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570   ∈ wcel 2145  ∃*wmo 2563  ∃*wrmo 3365  Disj wdisj 5070

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 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-mo 2565  df-cleq 2753  df-clel 2836  df-ral 3078  df-rmo 3366  df-ss 3916  df-disj 5071

This theorem is used by: (None)