Theorem sb5f 2040

Description: Equivalence for substitution when y is not free in ph. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6f.1	|- F/yph

Assertion

Ref	Expression
sb5f	|- ([y / x]ph <-> E.x(x = y /\ ph))

Proof of Theorem sb5f

Step	Hyp	Ref	Expression
1		sb6f.1	. . 3 |- F/yph
2	1	sb6f 2039	. 2 |- ([y / x]ph <-> A.x(x = y -> ph))
3	1	equs45f 1989	. 2 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))
4	2, 3	bitr4i 243	1 |- ([y / x]ph <-> E.x(x = y /\ ph))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358  A.wal 1540  E.wex 1541   F/wnf 1544  [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: (None)