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Theorem intabssd 44613
Description: When for each element 𝑦 there is a subset 𝐴 which may substituted for 𝑥 such that 𝑦 satisfying 𝜒 implies 𝑥 satisfies 𝜓 then the intersection of all 𝑥 that satisfy 𝜓 is a subclass the intersection of all 𝑦 that satisfy 𝜒. (Contributed by RP, 17-Oct-2020.)
Hypotheses
Ref Expression
intabssd.ex (𝜑 → 𝐴 ∈ 𝑉)
intabssd.sub ((𝜑 ∧ 𝑥 = 𝐴) → (𝜒 → 𝜓))
intabssd.ss (𝜑 → 𝐴 ⊆ 𝑦)
Assertion
Ref Expression
intabssd (𝜑 → ∩ {𝑥 ∣ 𝜓} ⊆ ∩ {𝑦 ∣ 𝜒})
Distinct variable groups:   𝜒,𝑥   𝜓,𝑦   𝑥,𝑦,𝜑   𝑥,𝐴
Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑦)   𝐴(𝑦)   𝑉(𝑥, 𝑦)

Proof of Theorem intabssd
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 intabssd.ex . . . . 5 (𝜑 → 𝐴 ∈ 𝑉)
2 intabssd.sub . . . . . 6 ((𝜑 ∧ 𝑥 = 𝐴) → (𝜒 → 𝜓))
3 eleq2 2849 . . . . . . . 8 (𝑥 = 𝐴 → (𝑧 ∈ 𝑥 ↔ 𝑧 ∈ 𝐴))
43biimpd 232 . . . . . . 7 (𝑥 = 𝐴 → (𝑧 ∈ 𝑥 → 𝑧 ∈ 𝐴))
5 intabssd.ss . . . . . . . 8 (𝜑 → 𝐴 ⊆ 𝑦)
65sseld 3907 . . . . . . 7 (𝜑 → (𝑧 ∈ 𝐴 → 𝑧 ∈ 𝑦))
74, 6sylan9r 518 . . . . . 6 ((𝜑 ∧ 𝑥 = 𝐴) → (𝑧 ∈ 𝑥 → 𝑧 ∈ 𝑦))
82, 7imim12d 82 . . . . 5 ((𝜑 ∧ 𝑥 = 𝐴) → ((𝜓 → 𝑧 ∈ 𝑥) → (𝜒 → 𝑧 ∈ 𝑦)))
91, 8spcimdv 3547 . . . 4 (𝜑 → (∀𝑥(𝜓 → 𝑧 ∈ 𝑥) → (𝜒 → 𝑧 ∈ 𝑦)))
109alrimdv 1961 . . 3 (𝜑 → (∀𝑥(𝜓 → 𝑧 ∈ 𝑥) → ∀𝑦(𝜒 → 𝑧 ∈ 𝑦)))
11 vex 3454 . . . 4 𝑧 ∈ V
1211elintab 4918 . . 3 (𝑧 ∈ ∩ {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜓 → 𝑧 ∈ 𝑥))
1311elintab 4918 . . 3 (𝑧 ∈ ∩ {𝑦 ∣ 𝜒} ↔ ∀𝑦(𝜒 → 𝑧 ∈ 𝑦))
1410, 12, 133imtr4g 299 . 2 (𝜑 → (𝑧 ∈ ∩ {𝑥 ∣ 𝜓} → 𝑧 ∈ ∩ {𝑦 ∣ 𝜒}))
1514ssrdv 3914 1 (𝜑 → ∩ {𝑥 ∣ 𝜓} ⊆ ∩ {𝑦 ∣ 𝜒})
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wal 1567   = wceq 1569   ∈ wcel 2144  {cab 2738   ⊆ wss 3895  ∩ cint 4906
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-v 3452  df-ss 3896  df-int 4907
This theorem is used by:  harval3  44632  clcnvlem  44716
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