Theorem sb56 2098

Description: Two equivalent ways of expressing the proper substitution of y for x in ph, when x and y are distinct. Theorem 6.2 of [Quine] p. 40. The proof does not involve df-sb 1649. (Contributed by NM, 14-Apr-2008.)

Assertion

Ref	Expression
sb56	|- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Distinct variable group:   x,y

Allowed substitution hints:   ph(x,y)

Proof of Theorem sb56

Step	Hyp	Ref	Expression
1		nfa1 1788	. 2 |- F/xA.x(x = y -> ph)
2		ax11v 2096	. . 3 |- (x = y -> (ph -> A.x(x = y -> ph)))
3		sp 1747	. . . 4 |- (A.x(x = y -> ph) -> (x = y -> ph))
4	3	com12 27	. . 3 |- (x = y -> (A.x(x = y -> ph) -> ph))
5	2, 4	impbid 183	. 2 |- (x = y -> (ph <-> A.x(x = y -> ph)))
6	1, 5	equsex 1962	1 |- (E.x(x = y /\ ph) <-> A.x(x = y -> ph))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358  A.wal 1540  E.wex 1541

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

This theorem is referenced by:  sb6  2099  sb5  2100  alexeq  2969