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Theorem k0004lem1 45091
Description: Application of ssin 4183 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 6657 . . . . . . 7 (𝐹 Fn 𝐴 → (𝐹𝐴) = ran 𝐹)
21sseq1d 3961 . . . . . 6 (𝐹 Fn 𝐴 → ((𝐹𝐴) ⊆ 𝐶 ↔ ran 𝐹𝐶))
32anbi2d 642 . . . . 5 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ (ran 𝐹𝐵 ∧ ran 𝐹𝐶)))
4 ssin 4183 . . . . 5 ((ran 𝐹𝐵 ∧ ran 𝐹𝐶) ↔ ran 𝐹 ⊆ (𝐵𝐶))
53, 4bitrdi 290 . . . 4 (𝐹 Fn 𝐴 → ((ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ ran 𝐹 ⊆ (𝐵𝐶)))
65pm5.32i 585 . . 3 ((𝐹 Fn 𝐴 ∧ (ran 𝐹𝐵 ∧ (𝐹𝐴) ⊆ 𝐶)) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
7 df-f 6531 . . . . 5 (𝐹:𝐴𝐵 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹𝐵))
87anbi1i 636 . . . 4 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ ((𝐹 Fn 𝐴 ∧ ran 𝐹𝐵) ∧ (𝐹𝐴) ⊆ 𝐶))
9 anass 474 . . . 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 6531 . . 3 (𝐹:𝐴⟶(𝐵𝐶) ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 ⊆ (𝐵𝐶)))
126, 10, 113bitr4i 306 . 2 ((𝐹:𝐴𝐵 ∧ (𝐹𝐴) ⊆ 𝐶) ↔ 𝐹:𝐴⟶(𝐵𝐶))
13 feq3 6677 . 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
This proof depends on syntax axioms:  wi 4  wb 209  wa 401   = wceq 1570  cin 3897  wss 3898  ran crn 5648  cima 5650   Fn wfn 6522  wf 6523
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 2732  ax-sep 5248  ax-pr 5390
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-br 5103  df-opab 5167  df-xp 5653  df-rel 5654  df-cnv 5655  df-dm 5657  df-rn 5658  df-res 5659  df-ima 5660  df-fun 6529  df-fn 6530  df-f 6531
This theorem is used by:  k0004lem2  45092
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