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Theorem fabexd 7956
Description: Existence of a set of functions. In contrast to fabex 7958 or fabexg 7957, the condition in the class abstraction does not contain the function explicitly, but the function can be derived from it. Therefore, this theorem is also applicable for more special functions like one-to-one, onto or one-to-one onto functions. (Contributed by AV, 20-May-2025.)
Hypotheses
Ref Expression
fabexd.f ((𝜑 ∧ 𝜓) → 𝑓:𝑋⟶𝑌)
fabexd.x (𝜑 → 𝑋 ∈ 𝑉)
fabexd.y (𝜑 → 𝑌 ∈ 𝑊)
Assertion
Ref Expression
fabexd (𝜑 → {𝑓 ∣ 𝜓} ∈ V)
Distinct variable groups:   𝑓,𝑋   𝑓,𝑌   𝜑,𝑓
Allowed substitution hints:   𝜓(𝑓)   𝑉(𝑓)   𝑊(𝑓)

Proof of Theorem fabexd
StepHypRef Expression
1 fabexd.x . . . 4 (𝜑 → 𝑋 ∈ 𝑉)
2 fabexd.y . . . 4 (𝜑 → 𝑌 ∈ 𝑊)
31, 2xpexd 7772 . . 3 (𝜑 → (𝑋 × 𝑌) ∈ V)
43pwexd 5344 . 2 (𝜑 → 𝒫 (𝑋 × 𝑌) ∈ V)
5 fabexd.f . . . . 5 ((𝜑 ∧ 𝜓) → 𝑓:𝑋⟶𝑌)
6 fssxp 6745 . . . . . 6 (𝑓:𝑋⟶𝑌 → 𝑓 ⊆ (𝑋 × 𝑌))
7 velpw 4564 . . . . . 6 (𝑓 ∈ 𝒫 (𝑋 × 𝑌) ↔ 𝑓 ⊆ (𝑋 × 𝑌))
86, 7sylibr 237 . . . . 5 (𝑓:𝑋⟶𝑌 → 𝑓 ∈ 𝒫 (𝑋 × 𝑌))
95, 8syl 18 . . . 4 ((𝜑 ∧ 𝜓) → 𝑓 ∈ 𝒫 (𝑋 × 𝑌))
109ex 418 . . 3 (𝜑 → (𝜓 → 𝑓 ∈ 𝒫 (𝑋 × 𝑌)))
1110abssdv 3994 . 2 (𝜑 → {𝑓 ∣ 𝜓} ⊆ 𝒫 (𝑋 × 𝑌))
124, 11ssexd 5289 1 (𝜑 → {𝑓 ∣ 𝜓} ∈ V)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∈ wcel 2146  {cab 2741  Vcvv 3453   ⊆ wss 3899  𝒫 cpw 4559   × cxp 5652  ⟶wf 6542
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735  ax-sep 5251  ax-pow 5330  ax-pr 5394  ax-un 7758
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3416  df-v 3455  df-ss 3900  df-dif 4049  df-in 4093  df-un 4153  df-nul 4281  df-if 4485  df-pw 4561  df-sn 4588  df-pr 4590  df-op 4838  df-uni 4870  df-br 5106  df-opab 5170  df-xp 5660  df-rel 5661  df-cnv 5662  df-dm 5664  df-rn 5665  df-fun 6548  df-fn 6549  df-f 6550
This theorem is used by:  fabexg  7957  f1oabexg  7960  grlimfn  49216  isgrlim  49219
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