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Theorem redundeq1 35851
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 3993 . . 3 (𝐴𝐵𝐷𝐵)
31ineq1i 4183 . . . 4 (𝐴𝐶) = (𝐷𝐶)
43eqeq1i 2824 . . 3 ((𝐴𝐶) = (𝐵𝐶) ↔ (𝐷𝐶) = (𝐵𝐶))
52, 4anbi12i 628 . 2 ((𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)) ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
6 df-redund 35846 . 2 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)))
7 df-redund 35846 . 2 (𝐷 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐷𝐵 ∧ (𝐷𝐶) = (𝐵𝐶)))
85, 6, 73bitr4i 305 1 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ 𝐷 Redund ⟨𝐵, 𝐶⟩)
Colors of variables: wff setvar class
Syntax hints:  wb 208  wa 398   = wceq 1530  cin 3933  wss 3934   Redund wredund 35461
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2153  ax-12 2169  ax-ext 2791
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1533  df-ex 1774  df-nf 1778  df-sb 2063  df-clab 2798  df-cleq 2812  df-clel 2891  df-rab 3145  df-in 3941  df-ss 3950  df-redund 35846
This theorem is referenced by:  refrelsredund3  35856
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