Theorem hbs1 2105

Description: x is not free in [y / x]ph when x and y are distinct. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
hbs1	|- ([y / x]ph -> A.x[y / x]ph)

Distinct variable group:   x,y

Allowed substitution hints:   ph(x,y)

Proof of Theorem hbs1

Step	Hyp	Ref	Expression
1		ax16 2045	. 2 |- (A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
2		hbsb2 2057	. 2 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
3	1, 2	pm2.61i 156	1 |- ([y / x]ph -> A.x[y / x]ph)

Syntax hints:   -> wi 4  A.wal 1540  [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:  nfs1v  2106  hbab1  2342