Theorem sbbib 2390

Description: Reversal of substitution. (Contributed by AV, 6-Aug-2023.) (Proof shortened by Wolf Lammen, 4-Sep-2023.)

Hypotheses

Ref	Expression
sbbib.y	⊢ Ⅎ𝑦𝜑
sbbib.x	⊢ Ⅎ𝑥𝜓

Assertion

Ref	Expression
sbbib	⊢ (∀𝑦([𝑦 / 𝑥]𝜑 ↔ 𝜓) ↔ ∀𝑥(𝜑 ↔ [𝑥 / 𝑦]𝜓))

Distinct variable group:   𝑥,𝑦

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

Proof of Theorem sbbib

Step	Hyp	Ref	Expression
1		nfs1v 2192	. . 3 ⊢ Ⅎ𝑥[𝑦 / 𝑥]𝜑
2		sbbib.x	. . 3 ⊢ Ⅎ𝑥𝜓
3	1, 2	nfbi 1935	. 2 ⊢ Ⅎ𝑥([𝑦 / 𝑥]𝜑 ↔ 𝜓)
4		sbbib.y	. . 3 ⊢ Ⅎ𝑦𝜑
5		nfs1v 2192	. . 3 ⊢ Ⅎ𝑦[𝑥 / 𝑦]𝜓
6	4, 5	nfbi 1935	. 2 ⊢ Ⅎ𝑦(𝜑 ↔ [𝑥 / 𝑦]𝜓)
7		sbequ12r 2287	. . 3 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
8		sbequ12 2286	. . 3 ⊢ (𝑦 = 𝑥 → (𝜓 ↔ [𝑥 / 𝑦]𝜓))
9	7, 8	bibi12d 348	. 2 ⊢ (𝑦 = 𝑥 → (([𝑦 / 𝑥]𝜑 ↔ 𝜓) ↔ (𝜑 ↔ [𝑥 / 𝑦]𝜓)))
10	3, 6, 9	cbvalv1 2370	1 ⊢ (∀𝑦([𝑦 / 𝑥]𝜑 ↔ 𝜓) ↔ ∀𝑥(𝜑 ↔ [𝑥 / 𝑦]𝜓))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1567  Ⅎwnf 1815  [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

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099

This theorem is used by:  sbbibvv  2391  dfich2  48633