Theorem sbcan 3089

Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016.)

Assertion

Ref	Expression
sbcan	⊢ ([̣A / x]̣(φ ∧ ψ) ↔ ([̣A / x]̣φ ∧ [̣A / x]̣ψ))

Proof of Theorem sbcan

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 3056	. 2 ⊢ ([̣A / x]̣(φ ∧ ψ) → A ∈ V)
2		sbcex 3056	. . 3 ⊢ ([̣A / x]̣ψ → A ∈ V)
3	2	adantl 452	. 2 ⊢ (([̣A / x]̣φ ∧ [̣A / x]̣ψ) → A ∈ V)
4		dfsbcq2 3050	. . 3 ⊢ (y = A → ([y / x](φ ∧ ψ) ↔ [̣A / x]̣(φ ∧ ψ)))
5		dfsbcq2 3050	. . . 4 ⊢ (y = A → ([y / x]φ ↔ [̣A / x]̣φ))
6		dfsbcq2 3050	. . . 4 ⊢ (y = A → ([y / x]ψ ↔ [̣A / x]̣ψ))
7	5, 6	anbi12d 691	. . 3 ⊢ (y = A → (([y / x]φ ∧ [y / x]ψ) ↔ ([̣A / x]̣φ ∧ [̣A / x]̣ψ)))
8		sban 2069	. . 3 ⊢ ([y / x](φ ∧ ψ) ↔ ([y / x]φ ∧ [y / x]ψ))
9	4, 7, 8	vtoclbg 2916	. 2 ⊢ (A ∈ V → ([̣A / x]̣(φ ∧ ψ) ↔ ([̣A / x]̣φ ∧ [̣A / x]̣ψ)))
10	1, 3, 9	pm5.21nii 342	1 ⊢ ([̣A / x]̣(φ ∧ ψ) ↔ ([̣A / x]̣φ ∧ [̣A / x]̣ψ))

Syntax hints:   ↔ wb 176   ∧ wa 358   = wceq 1642  [wsb 1648   ∈ wcel 1710  Vcvv 2860  [̣wsbc 3047

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-sbc 3048

This theorem is referenced by:  inopab  4863