Theorem disjeq1d 4000

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

Hypothesis

Ref	Expression
disjeq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
disjeq1d	⊢ (𝜑 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hints:   𝜑(𝑥)   𝐶(𝑥)

Proof of Theorem disjeq1d

Step	Hyp	Ref	Expression
1		disjeq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		disjeq1 3999	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))
3	1, 2	syl 14	1 ⊢ (𝜑 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1363  Disj wdisj 3992

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 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169

This theorem depends on definitions:  df-bi 117  df-nf 1471  df-sb 1773  df-eu 2039  df-mo 2040  df-clab 2174  df-cleq 2180  df-clel 2183  df-rmo 2473  df-in 3147  df-ss 3154  df-disj 3993

This theorem is referenced by:  disjeq12d  4001