Theorem sblbis 2072

Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993.)

Hypothesis

Ref	Expression
sblbis.1	⊢ ([y / x]φ ↔ ψ)

Assertion

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

Proof of Theorem sblbis

Step	Hyp	Ref	Expression
1		sbbi 2071	. 2 ⊢ ([y / x](χ ↔ φ) ↔ ([y / x]χ ↔ [y / x]φ))
2		sblbis.1	. . 3 ⊢ ([y / x]φ ↔ ψ)
3	2	bibi2i 304	. 2 ⊢ (([y / x]χ ↔ [y / x]φ) ↔ ([y / x]χ ↔ ψ))
4	1, 3	bitri 240	1 ⊢ ([y / x](χ ↔ φ) ↔ ([y / x]χ ↔ ψ))

Syntax hints:   ↔ wb 176  [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:  sb8eu  2222  sb8iota  4347