Theorem equs5 1591

Description: Lemma used in proofs of substitution properties.

Assertion

Ref	Expression
equs5	|- (-. A.x x = y -> (E.x(x = y /\ ph) -> A.x(x = y -> ph)))

Proof of Theorem equs5

Step	Hyp	Ref	Expression
1		hbnae 1507	. 2 |- (-. A.x x = y -> A.x -. A.x x = y)
2		hba1 1350	. 2 |- (A.x(x = y -> ph) -> A.xA.x(x = y -> ph))
3		ax-11o 1588	. . 3 |- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))
4	3	imp3a 388	. 2 |- (-. A.x x = y -> ((x = y /\ ph) -> A.x(x = y -> ph)))
5	1, 2, 4	19.23ad 1415	1 |- (-. A.x x = y -> (E.x(x = y /\ ph) -> A.x(x = y -> ph)))

Syntax hints:  -. wn 2   -> wi 3   /\ wa 240  A.wal 1296   = wceq 1298  E.wex 1326

This theorem is referenced by:  sb3 1592  sb4 1593

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-10 1308  ax-12 1310  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-10o 1500  ax-11o 1588

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327

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