Theorem sbbibvv 2396

Description: Reversal of substitution. (Contributed by AV, 6-Aug-2023.)

Assertion

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

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

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

Proof of Theorem sbbibvv

Step	Hyp	Ref	Expression
1		nfv 1947	. 2 ⊢ Ⅎ𝑦𝜑
2		nfv 1947	. 2 ⊢ Ⅎ𝑥𝜓
3	1, 2	sbbib 2395	1 ⊢ (∀𝑦([𝑦 / 𝑥]𝜑 ↔ 𝜓) ↔ ∀𝑥(𝜑 ↔ [𝑥 / 𝑦]𝜓))

This proof depends on syntax axioms:   ↔ 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

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by: (None)