Theorem sbequ 877

Description: An equality theorem for substitution. Used in proof of Theorem 9.7 in [Megill] p. 449 (p. 16 of the preprint).

Assertion

Ref	Expression
sbequ	|- (x = y -> ([x / z]ph <-> [y / z]ph))

Proof of Theorem sbequ

Step	Hyp	Ref	Expression
1		sbequi 876	. 2 |- (x = y -> ([x / z]ph -> [y / z]ph))
2		sbequi 876	. . 3 |- (y = x -> ([y / z]ph -> [x / z]ph))
3	2	eqcoms 813	. 2 |- (x = y -> ([y / z]ph -> [x / z]ph))
4	1, 3	impbid 397	1 |- (x = y -> ([x / z]ph <-> [y / z]ph))

Syntax hints:   -> wi 2   <-> wb 127   = weq 797  [wsb 852

This theorem is referenced by:  sbco2 913  findes 2400  tfinds 2401  tfindes 2404  nn1suc 4435

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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