Theorem absnw 43955

Description: Replace ax-10 2180, ax-11 2197, ax-12 2216 in absn 4604 with a substitution hypothesis. (Contributed by SN, 27-May-2025.)

Hypothesis

Ref	Expression
absnw.y	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
absnw	⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

Distinct variable groups:   𝜑,𝑦   𝜓,𝑥   𝑥,𝑦,𝑌

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem absnw

Step	Hyp	Ref	Expression
1		df-sn 4589	. . 3 ⊢ {𝑌} = {𝑥 ∣ 𝑥 = 𝑌}
2	1	eqeq2i 2776	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌})
3		absnw.y	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4		eqeq1 2767	. . 3 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑌 ↔ 𝑦 = 𝑌))
5	3, 4	abbibw 43954	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝑥 = 𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))
6	2, 5	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑌} ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑌))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570  {cab 2741  {csn 4588

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-sn 4589

This theorem is used by:  euabsn2w  43956