Theorem pprodeq1 5835

Description: Equality theorem for parallel product. (Contributed by Scott Fenton, 31-Jul-2019.)

Assertion

Ref	Expression
pprodeq1	|- (A = B -> PProd (A, C) = PProd (B, C))

Proof of Theorem pprodeq1

Step	Hyp	Ref	Expression
1		coeq1 4875	. . 3 |- (A = B -> (A o. 1st) = (B o. 1st))
2		txpeq1 5780	. . 3 |- ((A o. 1st) = (B o. 1st) -> ((A o. 1st) (x) (C o. 2nd)) = ((B o. 1st) (x) (C o. 2nd)))
3	1, 2	syl 15	. 2 |- (A = B -> ((A o. 1st) (x) (C o. 2nd)) = ((B o. 1st) (x) (C o. 2nd)))
4		df-pprod 5739	. 2 |- PProd (A, C) = ((A o. 1st) (x) (C o. 2nd))
5		df-pprod 5739	. 2 |- PProd (B, C) = ((B o. 1st) (x) (C o. 2nd))
6	3, 4, 5	3eqtr4g 2410	1 |- (A = B -> PProd (A, C) = PProd (B, C))

Syntax hints:   -> wi 4   = wceq 1642  1stc1st 4718   o. ccom 4722  2ndc2nd 4784   (x) ctxp 5736   PProd cpprod 5738

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-opab 4624  df-br 4641  df-co 4727  df-txp 5737  df-pprod 5739

This theorem is referenced by: (None)