Theorem sblbis 2013

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

Hypothesis

Ref	Expression
sblbis.1	|- ( [ y / x ] ph <-> ps )

Assertion

Ref	Expression
sblbis	|- ( [ y / x ] ( ch <-> ph ) <-> ( [ y / x ] ch <-> ps ) )

Proof of Theorem sblbis

Step	Hyp	Ref	Expression
1		sbbi 2012	. 2 |- ( [ y / x ] ( ch <-> ph ) <-> ( [ y / x ] ch <-> [ y / x ] ph ) )
2		sblbis.1	. . 3 |- ( [ y / x ] ph <-> ps )
3	2	bibi2i 227	. 2 |- ( ( [ y / x ] ch <-> [ y / x ] ph ) <-> ( [ y / x ] ch <-> ps ) )
4	1, 3	bitri 184	1 |- ( [ y / x ] ( ch <-> ph ) <-> ( [ y / x ] ch <-> ps ) )

Syntax hints:   <-> wb 105   [wsb 1810

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811

This theorem is referenced by:  sb8eu  2092  sb8euh  2102  sb8iota  5294