Theorem dffv4g 5692

Description: The previous definition of function value, from before the ℩ operator was introduced. Although based on the idea embodied by Definition 10.2 of [Quine] p. 65 (see args 5156), this definition apparently does not appear in the literature. (Contributed by NM, 1-Aug-1994.)

Assertion

Ref	Expression
dffv4g	⊢ (𝐴 ∈ 𝑉 → (𝐹‘𝐴) = ∪ {𝑥 ∣ (𝐹 “ {𝐴}) = {𝑥}})

Distinct variable groups:   𝑥,𝐴   𝑥,𝐹   𝑥,𝑉

Proof of Theorem dffv4g

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dffv3g 5691	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝐹‘𝐴) = (℩𝑦𝑦 ∈ (𝐹 “ {𝐴})))
2		df-iota 5337	. . 3 ⊢ (℩𝑦𝑦 ∈ (𝐹 “ {𝐴})) = ∪ {𝑥 ∣ {𝑦 ∣ 𝑦 ∈ (𝐹 “ {𝐴})} = {𝑥}}
3		abid2 2361	. . . . . 6 ⊢ {𝑦 ∣ 𝑦 ∈ (𝐹 “ {𝐴})} = (𝐹 “ {𝐴})
4	3	eqeq1i 2246	. . . . 5 ⊢ ({𝑦 ∣ 𝑦 ∈ (𝐹 “ {𝐴})} = {𝑥} ↔ (𝐹 “ {𝐴}) = {𝑥})
5	4	abbii 2354	. . . 4 ⊢ {𝑥 ∣ {𝑦 ∣ 𝑦 ∈ (𝐹 “ {𝐴})} = {𝑥}} = {𝑥 ∣ (𝐹 “ {𝐴}) = {𝑥}}
6	5	unieqi 3945	. . 3 ⊢ ∪ {𝑥 ∣ {𝑦 ∣ 𝑦 ∈ (𝐹 “ {𝐴})} = {𝑥}} = ∪ {𝑥 ∣ (𝐹 “ {𝐴}) = {𝑥}}
7	2, 6	eqtri 2259	. 2 ⊢ (℩𝑦𝑦 ∈ (𝐹 “ {𝐴})) = ∪ {𝑥 ∣ (𝐹 “ {𝐴}) = {𝑥}}
8	1, 7	eqtrdi 2287	1 ⊢ (𝐴 ∈ 𝑉 → (𝐹‘𝐴) = ∪ {𝑥 ∣ (𝐹 “ {𝐴}) = {𝑥}})

This proof depends on syntax axioms:   → wi 4   = wceq 1402   ∈ wcel 2209  {cab 2224  {csn 3709  ∪ cuni 3935   “ cima 4777  ℩cio 5335  ‘cfv 5377

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4249  ax-pow 4311  ax-pr 4346

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-sbc 3052  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-uni 3936  df-br 4131  df-opab 4193  df-xp 4780  df-cnv 4782  df-dm 4784  df-rn 4785  df-res 4786  df-ima 4787  df-iota 5337  df-fv 5385

This theorem is used by: (None)