Theorem abbib

Description: Equal class abstractions require equivalent formulas, and conversely. (Contributed by NM, 25-Nov-2013) (Revised by Mario Carneiro, 11-Aug-2016) Remove dependency on ax-8 and df-clel (by avoiding use of cleqh ). (Revised by BJ, 23-Jun-2019) Reverse the biconditional. (Revised by Wolf Lammen, 23-Feb-2025) (Proof shortened by SN, 22-Oct-2026)

	Ref	Expression
Assertion	abbib	⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑥 ∣ 𝜓 } ↔ ∀ 𝑥 ( 𝜑 ↔ 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
2		df-clab	⊢ ( 𝑦 ∈ { 𝑥 ∣ 𝜓 } ↔ [ 𝑦 / 𝑥 ] 𝜓 )
3	1 2	bibi12i	⊢ ( ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜓 ) )
4		sbbi	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜓 ) )
5	3 4	bitr4i	⊢ ( ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) )
6	5	albii	⊢ ( ∀ 𝑦 ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) ↔ ∀ 𝑦 [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) )
7		dfcleq	⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑥 ∣ 𝜓 } ↔ ∀ 𝑦 ( 𝑦 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ 𝜓 } ) )
8		sb8v	⊢ ( ∀ 𝑥 ( 𝜑 ↔ 𝜓 ) ↔ ∀ 𝑦 [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) )
9	6 7 8	3bitr4i	⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑥 ∣ 𝜓 } ↔ ∀ 𝑥 ( 𝜑 ↔ 𝜓 ) )