Theorem nfa1w 43467

Description: Replace ax-10 2179 in nfa1 2189 with a substitution hypothesis. (Contributed by SN, 2-Sep-2025.)

Hypothesis

Ref	Expression
nfa1w.x	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
nfa1w	⊢ Ⅎ𝑥∀𝑥𝜑

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

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

Proof of Theorem nfa1w

Step	Hyp	Ref	Expression
1		nfa1w.x	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	cbvalvw 2069	. 2 ⊢ (∀𝑥𝜑 ↔ ∀𝑦𝜓)
3		nfv 1947	. 2 ⊢ Ⅎ𝑥∀𝑦𝜓
4	2, 3	nfxfr 1886	1 ⊢ Ⅎ𝑥∀𝑥𝜑

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568  Ⅎwnf 1816

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 1943  ax-6 2000  ax-7 2041

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817

This theorem is used by:  eu6w  43468