Theorem rabeqdv 2627

Description: Equality of restricted class abstractions. Deduction form of rabeq 2625. (Contributed by Glauco Siliprandi, 5-Apr-2020.)

Hypothesis

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

Assertion

Ref	Expression
rabeqdv	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜑,𝑥

Allowed substitution hint:   𝜓(𝑥)

Proof of Theorem rabeqdv

Step	Hyp	Ref	Expression
1		rabeqdv.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		rabeq 2625	. 2 ⊢ (𝐴 = 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})
3	1, 2	syl 14	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Syntax hints:   → wi 4   = wceq 1296  {crab 2374

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077

This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-rab 2379

This theorem is referenced by:  dfphi2  11623  cncfval  12325