Theorem disjeq1 4108

Description: Equality theorem for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016.)

Assertion

Ref	Expression
disjeq1	⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝐶(𝑥)

Proof of Theorem disjeq1

Step	Hyp	Ref	Expression
1		eqimss2 3303	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		disjss1 4107	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶))
3	1, 2	syl 14	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 → Disj 𝑥 ∈ 𝐵 𝐶))
4		eqimss 3302	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		disjss1 4107	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))
6	4, 5	syl 14	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐵 𝐶 → Disj 𝑥 ∈ 𝐴 𝐶))
7	3, 6	impbid 129	1 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1402   ⊆ wss 3220  Disj wdisj 4101

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-rmo 2536  df-in 3226  df-ss 3233  df-disj 4102

This theorem is referenced by:  disjeq1d  4109