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Theorem redundeq1 36742
Description: Equivalence of redundancy predicates. (Contributed by Peter Mazsa, 26-Oct-2022.)
Hypothesis
Ref Expression
redundeq1.1 𝐴 = 𝐷
Assertion
Ref Expression
redundeq1 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)

Proof of Theorem redundeq1
StepHypRef Expression
1 redundeq1.1 . . . 4 𝐴 = 𝐷
21sseq1i 3949 . . 3 (𝐴𝐵𝐷𝐵)
31ineq1i 4142 . . . 4 (𝐴𝐶) = (𝐷𝐶)
43eqeq1i 2743 . . 3 ((𝐴𝐶) = (𝐵𝐶) ↔ (𝐷𝐶) = (𝐵𝐶))
52, 4anbi12i 627 . 2 ((𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)) ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
6 df-redund 36737 . 2 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)))
7 df-redund 36737 . 2 (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
85, 6, 73bitr4i 303 1 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)
Colors of variables: wff setvar class
Syntax hints:  wb 205  wa 396   = wceq 1539  cin 3886  wss 3887   Redund wredund 36354
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-in 3894  df-ss 3904  df-redund 36737
This theorem is referenced by:  refrelsredund3  36747
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