Metamath Proof Explorer


Theorem absnw

Description: Replace ax-10 , ax-11 , ax-12 in absn with a substitution hypothesis. (Contributed by SN, 27-May-2025)

Ref Expression
Hypothesis absnw.y ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion absnw ( { 𝑥 ∣ 𝜑 } = { 𝑌 } ↔ ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 absnw.y ⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
2 df-sn ⊢ { 𝑌 } = { 𝑥 ∣ 𝑥 = 𝑌 }
3 2 eqeq2i ⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑌 } ↔ { 𝑥 ∣ 𝜑 } = { 𝑥 ∣ 𝑥 = 𝑌 } )
4 eqeq1 ⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑌 ↔ 𝑦 = 𝑌 ) )
5 1 4 abbibw ⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑥 ∣ 𝑥 = 𝑌 } ↔ ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑌 ) )
6 3 5 bitri ⊢ ( { 𝑥 ∣ 𝜑 } = { 𝑌 } ↔ ∀ 𝑥 ( 𝜑 ↔ 𝑥 = 𝑌 ) )