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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]φ ↔ [y / z]φ))

Proof of Theorem sbequ
StepHypRef Expression
1 sbequi 876 . 2 (x = y → ([x / z]φ → [y / z]φ))
2 sbequi 876 . . 3 (y = x → ([y / z]φ → [x / z]φ))
32eqcoms 813 . 2 (x = y → ([y / z]φ → [x / z]φ))
41, 3impbid 397 1 (x = y → ([x / z]φ ↔ [y / z]φ))
Colors of variables: wff set class
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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