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Theorem redundeq1 36385
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 3905 . . 3 (𝐴𝐵𝐷𝐵)
31ineq1i 4099 . . . 4 (𝐴𝐶) = (𝐷𝐶)
43eqeq1i 2743 . . 3 ((𝐴𝐶) = (𝐵𝐶) ↔ (𝐷𝐶) = (𝐵𝐶))
52, 4anbi12i 630 . 2 ((𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)) ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
6 df-redund 36380 . 2 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)))
7 df-redund 36380 . 2 (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
85, 6, 73bitr4i 306 1 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 399   = wceq 1542  cin 3842  wss 3843   Redund wredund 35997
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-ext 2710
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-ex 1787  df-sb 2075  df-clab 2717  df-cleq 2730  df-clel 2811  df-rab 3062  df-v 3400  df-in 3850  df-ss 3860  df-redund 36380
This theorem is referenced by:  refrelsredund3  36390
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