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Theorem p6eq 4238
Description: Equality theorem for P6 operation. (Contributed by SF, 12-Jan-2015.)
Assertion
Ref Expression
p6eq (A = BP6 A = P6 B)

Proof of Theorem p6eq
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 sseq2 3293 . . 3 (A = B → ((V ×k {{x}}) A ↔ (V ×k {{x}}) B))
21abbidv 2467 . 2 (A = B → {x (V ×k {{x}}) A} = {x (V ×k {{x}}) B})
3 df-p6 4191 . 2 P6 A = {x (V ×k {{x}}) A}
4 df-p6 4191 . 2 P6 B = {x (V ×k {{x}}) B}
52, 3, 43eqtr4g 2410 1 (A = BP6 A = P6 B)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1642  {cab 2339  Vcvv 2859   wss 3257  {csn 3737   ×k cxpk 4174   P6 cp6 4178
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  ax-ext 2334
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-ss 3259  df-p6 4191
This theorem is referenced by:  p6eqi  4239  p6eqd  4240  p6exg  4290
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