Theorem fundmeng 8979

Description: A function is equinumerous to its domain. Exercise 4 of [Suppes] p. 98. (Contributed by NM, 17-Sep-2013.)

Assertion

Ref	Expression
fundmeng	⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)

Proof of Theorem fundmeng

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		funeq 6518	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 5858	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 22	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 5098	. . . 4 ⊢ (𝑥 = 𝐹 → (dom 𝑥 ≈ 𝑥 ↔ dom 𝐹 ≈ 𝐹))
5	1, 4	imbi12d 344	. . 3 ⊢ (𝑥 = 𝐹 → ((Fun 𝑥 → dom 𝑥 ≈ 𝑥) ↔ (Fun 𝐹 → dom 𝐹 ≈ 𝐹)))
6		vex 3433	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 8978	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 3499	. 2 ⊢ (𝐹 ∈ 𝑉 → (Fun 𝐹 → dom 𝐹 ≈ 𝐹))
9	8	imp 406	1 ⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1542   ∈ wcel 2114   class class class wbr 5085  dom cdm 5631  Fun wfun 6492   ≈ cen 8890

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-10 2147  ax-11 2163  ax-12 2185  ax-ext 2708  ax-sep 5231  ax-nul 5241  ax-pow 5307  ax-pr 5375  ax-un 7689

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-nf 1786  df-sb 2069  df-mo 2539  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2811  df-nfc 2885  df-ne 2933  df-ral 3052  df-rex 3062  df-rab 3390  df-v 3431  df-dif 3892  df-un 3894  df-in 3896  df-ss 3906  df-nul 4274  df-if 4467  df-pw 4543  df-sn 4568  df-pr 4570  df-op 4574  df-uni 4851  df-int 4890  df-br 5086  df-opab 5148  df-mpt 5167  df-id 5526  df-xp 5637  df-rel 5638  df-cnv 5639  df-co 5640  df-dm 5641  df-rn 5642  df-iota 6454  df-fun 6500  df-fn 6501  df-f 6502  df-f1 6503  df-fo 6504  df-f1o 6505  df-fv 6506  df-en 8894

This theorem is referenced by:  fndmeng  8982