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 ∘ 1st ) = (B ∘ 1st ))
2		txpeq1 5780	. . 3 ⊢ ((A ∘ 1st ) = (B ∘ 1st ) → ((A ∘ 1st ) ⊗ (C ∘ 2nd )) = ((B ∘ 1st ) ⊗ (C ∘ 2nd )))
3	1, 2	syl 15	. 2 ⊢ (A = B → ((A ∘ 1st ) ⊗ (C ∘ 2nd )) = ((B ∘ 1st ) ⊗ (C ∘ 2nd )))
4		df-pprod 5739	. 2 ⊢ PProd (A, C) = ((A ∘ 1st ) ⊗ (C ∘ 2nd ))
5		df-pprod 5739	. 2 ⊢ PProd (B, C) = ((B ∘ 1st ) ⊗ (C ∘ 2nd ))
6	3, 4, 5	3eqtr4g 2410	1 ⊢ (A = B → PProd (A, C) = PProd (B, C))

This proof depends on syntax axioms:   → wi 4   = wceq 1642  1^st c1st 4718   ∘ ccom 4722  2^nd c2nd 4784   ⊗ ctxp 5736   PProd cpprod 5738

This proof depends on 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 proof 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 used by: (None)