Theorem sbhb 2555

Description: Two ways of expressing "𝑥 is (effectively) not free in 𝜑". Usage of this theorem is discouraged because it depends on ax-13 2406. (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 1947	. . . 4 ⊢ Ⅎ𝑦𝜑
2	1	sb8 2551	. . 3 ⊢ (∀𝑥𝜑 ↔ ∀𝑦[𝑦 / 𝑥]𝜑)
3	2	imbi2i 339	. 2 ⊢ ((𝜑 → ∀𝑥𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
4		19.21v 1972	. 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 1568  [wsb 2099

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  ax-10 2179  ax-11 2195  ax-12 2216  ax-13 2406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by: (None)