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Theorem mul4r 11338
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 11145 . . . 4 ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
21adantl 484 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
32oveq2d 7397 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐵) · (𝐷 · 𝐶)))
4 mul4 11337 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
54ancom2s 658 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
6 simplr 776 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐵 ∈ ℂ)
7 simprl 778 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐶 ∈ ℂ)
86, 7mulcomd 11189 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐵 · 𝐶) = (𝐶 · 𝐵))
98oveq2d 7397 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐷) · (𝐵 · 𝐶)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
103, 5, 93eqtrd 2791 1 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 398   = wceq 1550  wcel 2132  (class class class)co 7381  cc 11057   · cmul 11064
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1805  ax-4 1819  ax-5 1920  ax-6 1977  ax-7 2018  ax-8 2134  ax-9 2142  ax-ext 2724  ax-mulcl 11121  ax-mulcom 11123  ax-mulass 11125
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 857  df-3an 1097  df-tru 1553  df-fal 1563  df-ex 1790  df-sb 2081  df-clab 2731  df-cleq 2744  df-clel 2827  df-rab 3405  df-v 3446  df-dif 3898  df-un 3900  df-ss 3912  df-nul 4277  df-if 4471  df-sn 4573  df-pr 4575  df-op 4579  df-uni 4856  df-br 5091  df-iota 6462  df-fv 6514  df-ov 7384
This theorem is referenced by:  bhmafibid1cn  15465  bhmafibid2cn  15466  2itscplem2  49339
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