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Theorem mul4r 11479
Description: Rearrangement of 4 factors: swap the right factors in the factors of a product of two products. (Contributed by AV, 4-Mar-2023.)
Assertion
Ref Expression
mul4r (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))

Proof of Theorem mul4r
StepHypRef Expression
1 mulcom 11286 . . . 4 ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
21adantl 487 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
32oveq2d 7436 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐵) · (𝐷 · 𝐶)))
4 mul4 11478 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
54ancom2s 663 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
6 simplr 781 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐵 ∈ ℂ)
7 simprl 783 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐶 ∈ ℂ)
86, 7mulcomd 11330 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐵 · 𝐶) = (𝐶 · 𝐵))
98oveq2d 7436 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐷) · (𝐵 · 𝐶)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
103, 5, 93eqtrd 2799 1 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ∈ wcel 2144  (class class class)co 7420  ℂcc 11198   · cmul 11205
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-mulcl 11262  ax-mulcom 11264  ax-mulass 11266
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-un 4151  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5103  df-iota 6494  df-fv 6546  df-ov 7423
This theorem is used by:  bhmafibid1cn  15633  bhmafibid2cn  15634  2itscplem2  49934
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