Theorem sb3an 2070

Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)

Assertion

Ref	Expression
sb3an	⊢ ([y / x](φ ∧ ψ ∧ χ) ↔ ([y / x]φ ∧ [y / x]ψ ∧ [y / x]χ))

Proof of Theorem sb3an

Step	Hyp	Ref	Expression
1		df-3an 936	. . 3 ⊢ ((φ ∧ ψ ∧ χ) ↔ ((φ ∧ ψ) ∧ χ))
2	1	sbbii 1653	. 2 ⊢ ([y / x](φ ∧ ψ ∧ χ) ↔ [y / x]((φ ∧ ψ) ∧ χ))
3		sban 2069	. 2 ⊢ ([y / x]((φ ∧ ψ) ∧ χ) ↔ ([y / x](φ ∧ ψ) ∧ [y / x]χ))
4		sban 2069	. . . 4 ⊢ ([y / x](φ ∧ ψ) ↔ ([y / x]φ ∧ [y / x]ψ))
5	4	anbi1i 676	. . 3 ⊢ (([y / x](φ ∧ ψ) ∧ [y / x]χ) ↔ (([y / x]φ ∧ [y / x]ψ) ∧ [y / x]χ))
6		df-3an 936	. . 3 ⊢ (([y / x]φ ∧ [y / x]ψ ∧ [y / x]χ) ↔ (([y / x]φ ∧ [y / x]ψ) ∧ [y / x]χ))
7	5, 6	bitr4i 243	. 2 ⊢ (([y / x](φ ∧ ψ) ∧ [y / x]χ) ↔ ([y / x]φ ∧ [y / x]ψ ∧ [y / x]χ))
8	2, 3, 7	3bitri 262	1 ⊢ ([y / x](φ ∧ ψ ∧ χ) ↔ ([y / x]φ ∧ [y / x]ψ ∧ [y / x]χ))

This proof depends on syntax axioms:   ↔ wb 176   ∧ wa 358   ∧ w3a 934  [wsb 1648

This proof depends on 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 proof depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by:  sbc3ang  3105