Theorem sbbi 2071

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

Assertion

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

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		dfbi2 609	. . 3 ⊢ ((φ ↔ ψ) ↔ ((φ → ψ) ∧ (ψ → φ)))
2	1	sbbii 1653	. 2 ⊢ ([y / x](φ ↔ ψ) ↔ [y / x]((φ → ψ) ∧ (ψ → φ)))
3		sbim 2065	. . . 4 ⊢ ([y / x](φ → ψ) ↔ ([y / x]φ → [y / x]ψ))
4		sbim 2065	. . . 4 ⊢ ([y / x](ψ → φ) ↔ ([y / x]ψ → [y / x]φ))
5	3, 4	anbi12i 678	. . 3 ⊢ (([y / x](φ → ψ) ∧ [y / x](ψ → φ)) ↔ (([y / x]φ → [y / x]ψ) ∧ ([y / x]ψ → [y / x]φ)))
6		sban 2069	. . 3 ⊢ ([y / x]((φ → ψ) ∧ (ψ → φ)) ↔ ([y / x](φ → ψ) ∧ [y / x](ψ → φ)))
7		dfbi2 609	. . 3 ⊢ (([y / x]φ ↔ [y / x]ψ) ↔ (([y / x]φ → [y / x]ψ) ∧ ([y / x]ψ → [y / x]φ)))
8	5, 6, 7	3bitr4i 268	. 2 ⊢ ([y / x]((φ → ψ) ∧ (ψ → φ)) ↔ ([y / x]φ ↔ [y / x]ψ))
9	2, 8	bitri 240	1 ⊢ ([y / x](φ ↔ ψ) ↔ ([y / x]φ ↔ [y / x]ψ))

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

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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:  sblbis  2072  sbrbis  2073  spsbbi  2077  sbco  2083  sbidm  2085  sbal  2127  sb8eu  2222  pm13.183  2979  sbcbig  3092  sb8iota  4346