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Theorem k0004lem1 44799
Description: Application of ssin 4199 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 6666 . . . . . . 7 (𝐹 Fn 𝐴 → (𝐹𝐴) = ran 𝐹)
21sseq1d 3976 . . . . . 6 (𝐹 Fn 𝐴 → ((𝐹𝐴) ⊆ 𝐶 ↔ ran 𝐹𝐶))
32anbi2d 641 . . . . 5 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (ran 𝐹𝐵 ∧ ran 𝐹𝐶)))
4 ssin 4199 . . . . 5 ((ran 𝐹𝐵 ∧ ran 𝐹𝐶) ↔ ran 𝐹 ⊆ (𝐵𝐶))
53, 4bitrdi 290 . . . 4 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ ran 𝐹 ⊆ (𝐵𝐶)))
65pm5.32i 584 . . 3 ((𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
7 df-f 6541 . . . . 5 (𝐹:𝐴𝐵 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹𝐵))
87anbi1i 635 . . . 4 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ ((𝐹 Fn 𝐴 ∧ ran 𝐹𝐵) ∧ (𝐹𝐴) ⊆ 𝐶))
9 anass 473 . . . 4 (((𝐹 Fn 𝐴 ∧ ran 𝐹𝐵) ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)))
108, 9bitri 278 . . 3 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)))
11 df-f 6541 . . 3 (𝐹:𝐴⟶(𝐵𝐶) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
126, 10, 113bitr4i 306 . 2 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴⟶(𝐵𝐶))
13 feq3 6686 . 2 (𝐷 = (𝐵𝐶) → (𝐹:𝐴𝐷𝐹:𝐴⟶(𝐵𝐶)))
1412, 13bitr4id 293 1 (𝐷 = (𝐵𝐶) → ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴𝐷))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 400   = wceq 1567  cin 3912  wss 3913  ran crn 5663  cima 5665   Fn wfn 6532  wf 6533
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741  ax-sep 5261  ax-pr 5405
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rex 3096  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5114  df-opab 5178  df-xp 5668  df-rel 5669  df-cnv 5670  df-dm 5672  df-rn 5673  df-res 5674  df-ima 5675  df-fun 6539  df-fn 6540  df-f 6541
This theorem is referenced by:  k0004lem2  44800
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