Theorem disjeq1 3960

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 3192	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		disjss1 3959	. . 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 3191	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		disjss1 3959	. . 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 128	1 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1342   ⊆ wss 3111  Disj wdisj 3953

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-10 1492  ax-11 1493  ax-i12 1494  ax-bndl 1496  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146

This theorem depends on definitions:  df-bi 116  df-nf 1448  df-sb 1750  df-eu 2016  df-mo 2017  df-clab 2151  df-cleq 2157  df-clel 2160  df-rmo 2450  df-in 3117  df-ss 3124  df-disj 3954

This theorem is referenced by:  disjeq1d  3961