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Theorem k0004lem1 44239
Description: Application of ssin 4186 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 6611 . . . . . . 7 (𝐹 Fn 𝐴 → (𝐹𝐴) = ran 𝐹)
21sseq1d 3961 . . . . . 6 (𝐹 Fn 𝐴 → ((𝐹𝐴) ⊆ 𝐶 ↔ ran 𝐹𝐶))
32anbi2d 630 . . . . 5 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (ran 𝐹𝐵 ∧ ran 𝐹𝐶)))
4 ssin 4186 . . . . 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 6485 . . . . 5 (𝐹:𝐴𝐵 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹𝐵))
87anbi1i 624 . . . 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 6485 . . 3 (𝐹:𝐴⟶(𝐵𝐶) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
126, 10, 113bitr4i 303 . 2 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴⟶(𝐵𝐶))
13 feq3 6631 . 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 1541  cin 3896  wss 3897  ran crn 5615  cima 5617   Fn wfn 6476  wf 6477
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703  ax-sep 5232  ax-nul 5242  ax-pr 5368
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4281  df-if 4473  df-sn 4574  df-pr 4576  df-op 4580  df-br 5090  df-opab 5152  df-xp 5620  df-rel 5621  df-cnv 5622  df-dm 5624  df-rn 5625  df-res 5626  df-ima 5627  df-fun 6483  df-fn 6484  df-f 6485
This theorem is referenced by:  k0004lem2  44240
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