Theorem nceqd 6110

Description: Equality deduction for cardinality. (Contributed by SF, 24-Feb-2015.)

Hypothesis

Ref	Expression
nceqd.1	|- (ph -> A = B)

Assertion

Ref	Expression
nceqd	|- (ph -> Nc A = Nc B)

Proof of Theorem nceqd

Step	Hyp	Ref	Expression
1		nceqd.1	. 2 |- (ph -> A = B)
2		nceq 6108	. 2 |- (A = B -> Nc A = Nc B)
3	1, 2	syl 15	1 |- (ph -> Nc A = Nc B)

Syntax hints:   -> wi 4   = wceq 1642   Nc cnc 6091

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-rex 2620  df-sn 3741  df-ima 4727  df-ec 5947  df-nc 6101

This theorem is referenced by:  df1c3g  6141  ncspw1eu  6159  pw1eltc  6162  dflec2  6210  nchoicelem7  6293  nchoicelem9  6295  nchoicelem11  6297  nchoicelem12  6298  nchoicelem14  6300  nchoicelem15  6301  nchoicelem16  6302  nchoicelem17  6303  nchoice  6306