Theorem absnw 43829

Description: Replace ax-10 2179, ax-11 2196, ax-12 2215 in absn 4603 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 4588	. . 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 43828	. 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 1567   = wceq 1569  {cab 2741  {csn 4587

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3455  df-sn 4588

This theorem is used by:  euabsn2w  43830