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Theorem k0004lem1 44109
Description: Application of ssin 4260 to range of a function. (Contributed by RP, 1-Apr-2021.)
Assertion
Ref Expression
k0004lem1 (𝐷 = (𝐵𝐶) → ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴𝐷))

Proof of Theorem k0004lem1
StepHypRef Expression
1 fnima 6710 . . . . . . 7 (𝐹 Fn 𝐴 → (𝐹𝐴) = ran 𝐹)
21sseq1d 4040 . . . . . 6 (𝐹 Fn 𝐴 → ((𝐹𝐴) ⊆ 𝐶 ↔ ran 𝐹𝐶))
32anbi2d 629 . . . . 5 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (ran 𝐹𝐵 ∧ ran 𝐹𝐶)))
4 ssin 4260 . . . . 5 ((ran 𝐹𝐵 ∧ ran 𝐹𝐶) ↔ ran 𝐹 ⊆ (𝐵𝐶))
53, 4bitrdi 287 . . . 4 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ ran 𝐹 ⊆ (𝐵𝐶)))
65pm5.32i 574 . . 3 ((𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
7 df-f 6577 . . . . 5 (𝐹:𝐴𝐵 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹𝐵))
87anbi1i 623 . . . 4 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ ((𝐹 Fn 𝐴 ∧ ran 𝐹𝐵) ∧ (𝐹𝐴) ⊆ 𝐶))
9 anass 468 . . . 4 (((𝐹 Fn 𝐴 ∧ ran 𝐹𝐵) ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)))
108, 9bitri 275 . . 3 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)))
11 df-f 6577 . . 3 (𝐹:𝐴⟶(𝐵𝐶) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
126, 10, 113bitr4i 303 . 2 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴⟶(𝐵𝐶))
13 feq3 6730 . 2 (𝐷 = (𝐵𝐶) → (𝐹:𝐴𝐷𝐹:𝐴⟶(𝐵𝐶)))
1412, 13bitr4id 290 1 (𝐷 = (𝐵𝐶) → ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴𝐷))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1537  cin 3975  wss 3976  ran crn 5701  cima 5703   Fn wfn 6568  wf 6569
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pr 5447
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-xp 5706  df-rel 5707  df-cnv 5708  df-dm 5710  df-rn 5711  df-res 5712  df-ima 5713  df-fun 6575  df-fn 6576  df-f 6577
This theorem is referenced by:  k0004lem2  44110
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