Theorem stdpc4 1168

Description: The specialization axiom of standard predicate calculus. It states that if a statement ph holds for all x, then it also holds for the specific case of y (properly) substituted for x. Translated to traditional notation, it can be read: "A.xph(x) -> ph(y), provided that y is free for x in ph(x)." Axiom 4 of [Mendelson] p. 69. See also a4sbc 1916 and ra4sbc 1968.

Assertion

Ref	Expression
stdpc4	|- (A.xph -> [y / x]ph)

Proof of Theorem stdpc4

Step	Hyp	Ref	Expression
1		ax-1 4	. . 3 |- (ph -> (x = y -> ph))
2	1	19.20i 968	. 2 |- (A.xph -> A.x(x = y -> ph))
3		sb2 1160	. 2 |- (A.x(x = y -> ph) -> [y / x]ph)
4	2, 3	syl 10	1 |- (A.xph -> [y / x]ph)

Syntax hints:   -> wi 3  A.wal 950  [wsbc 1153

This theorem is referenced by:  sbf 1169  hbs1f 1172  sbea4 1227  sbia4 1228  sbba4 1229  sb8 1245  sb9i 1247  a4sbc 1916  ra4sbc 1968  nd1 4861  nd2 4862

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-gen 955  ax-9 1102

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-sb 1155

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