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Theorem redundss3 35928
 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 4164 . . . 4 ((𝐴𝐶) = (𝐵𝐶) → ((𝐴𝐶) ∩ 𝐷) = ((𝐵𝐶) ∩ 𝐷))
2 redundss3.1 . . . . . . . 8 𝐷𝐶
3 dfss 3936 . . . . . . . 8 (𝐷𝐶𝐷 = (𝐷𝐶))
42, 3mpbi 233 . . . . . . 7 𝐷 = (𝐷𝐶)
5 incom 4161 . . . . . . 7 (𝐷𝐶) = (𝐶𝐷)
64, 5eqtri 2847 . . . . . 6 𝐷 = (𝐶𝐷)
76ineq2i 4169 . . . . 5 (𝐴𝐷) = (𝐴 ∩ (𝐶𝐷))
8 inass 4179 . . . . 5 ((𝐴𝐶) ∩ 𝐷) = (𝐴 ∩ (𝐶𝐷))
97, 8eqtr4i 2850 . . . 4 (𝐴𝐷) = ((𝐴𝐶) ∩ 𝐷)
106ineq2i 4169 . . . . 5 (𝐵𝐷) = (𝐵 ∩ (𝐶𝐷))
11 inass 4179 . . . . 5 ((𝐵𝐶) ∩ 𝐷) = (𝐵 ∩ (𝐶𝐷))
1210, 11eqtr4i 2850 . . . 4 (𝐵𝐷) = ((𝐵𝐶) ∩ 𝐷)
131, 9, 123eqtr4g 2884 . . 3 ((𝐴𝐶) = (𝐵𝐶) → (𝐴𝐷) = (𝐵𝐷))
1413anim2i 619 . 2 ((𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)) → (𝐴𝐵 ∧ (𝐴𝐷) = (𝐵𝐷)))
15 df-redund 35924 . 2 (𝐴 Redund ⟨𝐵, 𝐶⟩ ↔ (𝐴𝐵 ∧ (𝐴𝐶) = (𝐵𝐶)))
16 df-redund 35924 . 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 1538   ∩ cin 3917   ⊆ wss 3918   Redund wredund 35539 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 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796 This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-rab 3141  df-v 3481  df-in 3925  df-ss 3935  df-redund 35924 This theorem is referenced by:  refrelsredund2  35933
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