Theorem sbhb 2550

Description: Two ways of expressing "𝑥 is (effectively) not free in 𝜑". Usage of this theorem is discouraged because it depends on ax-13 2401. (Contributed by NM, 29-May-2009.) (New usage is discouraged.)

Assertion

Ref	Expression
sbhb	⊢ ((𝜑 → ∀𝑥𝜑) ↔ ∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑))

Distinct variable group:   𝜑,𝑦

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem sbhb

Step	Hyp	Ref	Expression
1		nfv 1946	. . . 4 ⊢ Ⅎ𝑦𝜑
2	1	sb8 2546	. . 3 ⊢ (∀𝑥𝜑 ↔ ∀𝑦[𝑦 / 𝑥]𝜑)
3	2	imbi2i 339	. 2 ⊢ ((𝜑 → ∀𝑥𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
4		19.21v 1971	. 2 ⊢ (∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
5	3, 4	bitr4i 281	1 ⊢ ((𝜑 → ∀𝑥𝜑) ↔ ∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  [wsb 2098

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 1942  ax-6 1999  ax-7 2040  ax-10 2177  ax-11 2193  ax-12 2212  ax-13 2401

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-nf 1816  df-sb 2099

This theorem is used by: (None)