Theorem disjeq1 4042

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 3256	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		disjss1 4041	. . 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 3255	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		disjss1 4041	. . 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 1373   ⊆ wss 3174  Disj wdisj 4035

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-eu 2058  df-mo 2059  df-clab 2194  df-cleq 2200  df-clel 2203  df-rmo 2494  df-in 3180  df-ss 3187  df-disj 4036

This theorem is referenced by:  disjeq1d  4043