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Theorem redundss3 36508
Description: Implication of redundancy predicate. (Contributed by Peter Mazsa, 26-Oct-2022.)
Hypothesis
Ref Expression
redundss3.1 𝐷𝐶
Assertion
Ref Expression
redundss3 (𝐴 Redund ⟨𝐵, 𝐶⟩ → 𝐴 Redund ⟨𝐵, 𝐷⟩)

Proof of Theorem redundss3
StepHypRef Expression
1 ineq1 4135 . . . 4 ((𝐴𝐶) = (𝐵𝐶) → ((𝐴𝐶) ∩ 𝐷) = ((𝐵𝐶) ∩ 𝐷))
2 redundss3.1 . . . . . . . 8 𝐷𝐶
3 dfss 3899 . . . . . . . 8 (𝐷𝐶𝐷 = (𝐷𝐶))
42, 3mpbi 233 . . . . . . 7 𝐷 = (𝐷𝐶)
5 incom 4130 . . . . . . 7 (𝐷𝐶) = (𝐶𝐷)
64, 5eqtri 2766 . . . . . 6 𝐷 = (𝐶𝐷)
76ineq2i 4139 . . . . 5 (𝐴𝐷) = (𝐴 ∩ (𝐶𝐷))
8 inass 4149 . . . . 5 ((𝐴𝐶) ∩ 𝐷) = (𝐴 ∩ (𝐶𝐷))
97, 8eqtr4i 2769 . . . 4 (𝐴𝐷) = ((𝐴𝐶) ∩ 𝐷)
106ineq2i 4139 . . . . 5 (𝐵𝐷) = (𝐵 ∩ (𝐶𝐷))
11 inass 4149 . . . . 5 ((𝐵𝐶) ∩ 𝐷) = (𝐵 ∩ (𝐶𝐷))
1210, 11eqtr4i 2769 . . . 4 (𝐵𝐷) = ((𝐵𝐶) ∩ 𝐷)
131, 9, 123eqtr4g 2804 . . 3 ((𝐴𝐶) = (𝐵𝐶) → (𝐴𝐷) = (𝐵𝐷))
1413anim2i 620 . 2 ((𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)) → (𝐴𝐵 ∧ (𝐴𝐷) = (𝐵𝐷)))
15 df-redund 36504 . 2 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)))
16 df-redund 36504 . 2 (𝐴 Redund ⟨𝐵, 𝐷⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐷) = (𝐵𝐷)))
1714, 15, 163imtr4i 295 1 (𝐴 Redund ⟨𝐵, 𝐶⟩ → 𝐴 Redund ⟨𝐵, 𝐷⟩)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 399   = wceq 1543  cin 3880  wss 3881   Redund wredund 36121
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2113  ax-9 2121  ax-ext 2709
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-ex 1788  df-sb 2072  df-clab 2716  df-cleq 2730  df-clel 2817  df-rab 3071  df-v 3423  df-in 3888  df-ss 3898  df-redund 36504
This theorem is referenced by:  refrelsredund2  36513
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