Theorem rabbida4 3436

Description: Version of rabbidva2 3414 with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by BJ, 27-Apr-2019.)

Hypotheses

Ref	Expression
rabbida4.nf	⊢ Ⅎ𝑥𝜑
rabbida4.1	⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐵 ∧ 𝜒)))

Assertion

Ref	Expression
rabbida4	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

Proof of Theorem rabbida4

Step	Hyp	Ref	Expression
1		rabbida4.nf	. . 3 ⊢ Ⅎ𝑥𝜑
2		rabbida4.1	. . 3 ⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝑥 ∈ 𝐵 ∧ 𝜒)))
3	1, 2	abbid 2828	. 2 ⊢ (𝜑 → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜒)})
4		df-rab 3413	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
5		df-rab 3413	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜒} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜒)}
6	3, 4, 5	3eqtr4g 2820	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569  Ⅎwnf 1815   ∈ wcel 2144  {cab 2738  {crab 3412

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-9 2154  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-rab 3413

This theorem is used by:  rabbida  3437  rabeqd  3439  rabeqf  3445