Metamath Proof Explorer


Theorem rabid2im

Description: One direction of rabid2 is based on fewer axioms. (Contributed by Wolf Lammen, 26-May-2025)

Ref Expression
Assertion rabid2im ( ∀ 𝑥 ∈ 𝐴 𝜑 → 𝐴 = { 𝑥 ∈ 𝐴 ∣ 𝜑 } )

Proof

Step Hyp Ref Expression
1 pm4.71 ⊢ ( ( 𝑥 ∈ 𝐴 → 𝜑 ) ↔ ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
2 1 albii ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
3 eqab ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) → 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } )
4 2 3 sylbi ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) → 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } )
5 df-ral ⊢ ( ∀ 𝑥 ∈ 𝐴 𝜑 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜑 ) )
6 df-rab ⊢ { 𝑥 ∈ 𝐴 ∣ 𝜑 } = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) }
7 6 eqeq2i ⊢ ( 𝐴 = { 𝑥 ∈ 𝐴 ∣ 𝜑 } ↔ 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) } )
8 4 5 7 3imtr4i ⊢ ( ∀ 𝑥 ∈ 𝐴 𝜑 → 𝐴 = { 𝑥 ∈ 𝐴 ∣ 𝜑 } )