Theorem p6eq 4239

Description: Equality theorem for P6 operation. (Contributed by SF, 12-Jan-2015.)

Assertion

Ref	Expression
p6eq	⊢ (A = B → P6 A = P6 B)

Proof of Theorem p6eq

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sseq2 3294	. . 3 ⊢ (A = B → ((V ×k {{x}}) ⊆ A ↔ (V ×k {{x}}) ⊆ B))
2	1	abbidv 2468	. 2 ⊢ (A = B → {x ∣ (V ×k {{x}}) ⊆ A} = {x ∣ (V ×k {{x}}) ⊆ B})
3		df-p6 4192	. 2 ⊢ P6 A = {x ∣ (V ×k {{x}}) ⊆ A}
4		df-p6 4192	. 2 ⊢ P6 B = {x ∣ (V ×k {{x}}) ⊆ B}
5	2, 3, 4	3eqtr4g 2410	1 ⊢ (A = B → P6 A = P6 B)

Syntax hints:   → wi 4   = wceq 1642  {cab 2339  Vcvv 2860   ⊆ wss 3258  {csn 3738   ×_k cxpk 4175   P6 cp6 4179

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 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260  df-p6 4192

This theorem is referenced by:  p6eqi  4240  p6eqd  4241  p6exg  4291