Theorem sbequ8 2079

Description: Elimination of equality from antecedent after substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbequ8	|- ([y / x]ph <-> [y / x](x = y -> ph))

Proof of Theorem sbequ8

Step	Hyp	Ref	Expression
1		equsb1 2034	. . 3 |- [y / x]x = y
2	1	a1bi 327	. 2 |- ([y / x]ph <-> ([y / x]x = y -> [y / x]ph))
3		sbim 2065	. 2 |- ([y / x](x = y -> ph) <-> ([y / x]x = y -> [y / x]ph))
4	2, 3	bitr4i 243	1 |- ([y / x]ph <-> [y / x](x = y -> ph))

Syntax hints:   -> wi 4   <-> 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: (None)