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Theorem k0004lem1 44532
Description: Application of ssin 4193 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 6632 . . . . . . 7 (𝐹 Fn 𝐴 → (𝐹𝐴) = ran 𝐹)
21sseq1d 3967 . . . . . 6 (𝐹 Fn 𝐴 → ((𝐹𝐴) ⊆ 𝐶 ↔ ran 𝐹𝐶))
32anbi2d 631 . . . . 5 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (ran 𝐹𝐵 ∧ ran 𝐹𝐶)))
4 ssin 4193 . . . . 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 6506 . . . . 5 (𝐹:𝐴𝐵 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹𝐵))
87anbi1i 625 . . . 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 6506 . . 3 (𝐹:𝐴⟶(𝐵𝐶) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
126, 10, 113bitr4i 303 . 2 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴⟶(𝐵𝐶))
13 feq3 6652 . 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 1542  cin 3902  wss 3903  ran crn 5635  cima 5637   Fn wfn 6497  wf 6498
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 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5245  ax-pr 5381
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-xp 5640  df-rel 5641  df-cnv 5642  df-dm 5644  df-rn 5645  df-res 5646  df-ima 5647  df-fun 6504  df-fn 6505  df-f 6506
This theorem is referenced by:  k0004lem2  44533
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