Theorem disjeq1d 5077

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 5076	. 2 ⊢ (𝐴 = 𝐵 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))
3	1, 2	syl 18	1 ⊢ (𝜑 → (Disj 𝑥 ∈ 𝐴 𝐶 ↔ Disj 𝑥 ∈ 𝐵 𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  Disj wdisj 5069

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-mo 2564  df-cleq 2752  df-clel 2835  df-rmo 3365  df-ss 3896  df-disj 5070

This theorem is used by:  disjeq12d  5078  disjxiun  5099  disjdifprg  33169  disjdifprg2  33170  disjun0  33189  tocyccntz  33705  measxun2  34843  measssd  34848  meadjun  47472