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Theorem setrecseq 50732
Description: Equality theorem for set recursion. (Contributed by Emmett Weisz, 17-Feb-2021.)
Assertion
Ref Expression
setrecseq (𝐹 = 𝐺 → setrecs(𝐹) = setrecs(𝐺))

Proof of Theorem setrecseq
Dummy variables 𝑦 𝑤 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fveq1 6876 . . . . . . . . . 10 (𝐹 = 𝐺 → (𝐹‘𝑤) = (𝐺‘𝑤))
21sseq1d 3962 . . . . . . . . 9 (𝐹 = 𝐺 → ((𝐹‘𝑤) ⊆ 𝑧 ↔ (𝐺‘𝑤) ⊆ 𝑧))
32imbi2d 343 . . . . . . . 8 (𝐹 = 𝐺 → ((𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧) ↔ (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧)))
43imbi2d 343 . . . . . . 7 (𝐹 = 𝐺 → ((𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) ↔ (𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧))))
54albidv 1953 . . . . . 6 (𝐹 = 𝐺 → (∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) ↔ ∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧))))
65imbi1d 344 . . . . 5 (𝐹 = 𝐺 → ((∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧) ↔ (∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)))
76albidv 1953 . . . 4 (𝐹 = 𝐺 → (∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧) ↔ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)))
87abbidv 2827 . . 3 (𝐹 = 𝐺 → {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)} = {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)})
98unieqd 4880 . 2 (𝐹 = 𝐺 → ∪ {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)} = ∪ {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)})
10 df-setrecs 9946 . 2 setrecs(𝐹) = ∪ {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐹‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)}
11 df-setrecs 9946 . 2 setrecs(𝐺) = ∪ {𝑦 ∣ ∀𝑧(∀𝑤(𝑤 ⊆ 𝑦 → (𝑤 ⊆ 𝑧 → (𝐺‘𝑤) ⊆ 𝑧)) → 𝑦 ⊆ 𝑧)}
129, 10, 113eqtr4g 2821 1 (𝐹 = 𝐺 → setrecs(𝐹) = setrecs(𝐺))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4  ∀wal 1568   = wceq 1570  {cab 2739   ⊆ wss 3899  ∪ cuni 4867  ‘cfv 6531  setrecscsetrecs 9945
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-ss 3916  df-uni 4868  df-br 5104  df-iota 6487  df-fv 6539  df-setrecs 9946
This theorem is used by: (None)
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