Theorem sban 2069

Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sban	⊢ ([y / x](φ ∧ ψ) ↔ ([y / x]φ ∧ [y / x]ψ))

Proof of Theorem sban

Step	Hyp	Ref	Expression
1		sbn 2062	. . 3 ⊢ ([y / x] ¬ (φ → ¬ ψ) ↔ ¬ [y / x](φ → ¬ ψ))
2		sbim 2065	. . . 4 ⊢ ([y / x](φ → ¬ ψ) ↔ ([y / x]φ → [y / x] ¬ ψ))
3		sbn 2062	. . . . 5 ⊢ ([y / x] ¬ ψ ↔ ¬ [y / x]ψ)
4	3	imbi2i 303	. . . 4 ⊢ (([y / x]φ → [y / x] ¬ ψ) ↔ ([y / x]φ → ¬ [y / x]ψ))
5	2, 4	bitri 240	. . 3 ⊢ ([y / x](φ → ¬ ψ) ↔ ([y / x]φ → ¬ [y / x]ψ))
6	1, 5	xchbinx 301	. 2 ⊢ ([y / x] ¬ (φ → ¬ ψ) ↔ ¬ ([y / x]φ → ¬ [y / x]ψ))
7		df-an 360	. . 3 ⊢ ((φ ∧ ψ) ↔ ¬ (φ → ¬ ψ))
8	7	sbbii 1653	. 2 ⊢ ([y / x](φ ∧ ψ) ↔ [y / x] ¬ (φ → ¬ ψ))
9		df-an 360	. 2 ⊢ (([y / x]φ ∧ [y / x]ψ) ↔ ¬ ([y / x]φ → ¬ [y / x]ψ))
10	6, 8, 9	3bitr4i 268	1 ⊢ ([y / x](φ ∧ ψ) ↔ ([y / x]φ ∧ [y / x]ψ))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358  [wsb 1648

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

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sb3an  2070  sbbi  2071  sbabel  2516  cbvreu  2834  sbcan  3089  sbcang  3090  rmo3  3134  inab  3523  difab  3524