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Theorem redundeq1 37487
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 4009 . . 3 (𝐴𝐵𝐷𝐵)
31ineq1i 4207 . . . 4 (𝐴𝐶) = (𝐷𝐶)
43eqeq1i 2737 . . 3 ((𝐴𝐶) = (𝐵𝐶) ↔ (𝐷𝐶) = (𝐵𝐶))
52, 4anbi12i 627 . 2 ((𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)) ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
6 df-redund 37482 . 2 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)))
7 df-redund 37482 . 2 (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
85, 6, 73bitr4i 302 1 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)
Colors of variables: wff setvar class
Syntax hints:  wb 205  wa 396   = wceq 1541  cin 3946  wss 3947   Redund wredund 37052
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-rab 3433  df-v 3476  df-in 3954  df-ss 3964  df-redund 37482
This theorem is referenced by:  refrelsredund3  37492
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