Theorem fundmeng 8980

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 6520	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 5857	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 22	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 5115	. . . 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 3448	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 8979	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 3517	. 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 1540   ∈ wcel 2109   class class class wbr 5102  dom cdm 5631  Fun wfun 6493   ≈ cen 8892

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2701  ax-sep 5246  ax-nul 5256  ax-pow 5315  ax-pr 5382  ax-un 7691

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2533  df-eu 2562  df-clab 2708  df-cleq 2721  df-clel 2803  df-nfc 2878  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3403  df-v 3446  df-dif 3914  df-un 3916  df-in 3918  df-ss 3928  df-nul 4293  df-if 4485  df-pw 4561  df-sn 4586  df-pr 4588  df-op 4592  df-uni 4868  df-int 4907  df-br 5103  df-opab 5165  df-mpt 5184  df-id 5526  df-xp 5637  df-rel 5638  df-cnv 5639  df-co 5640  df-dm 5641  df-rn 5642  df-iota 6452  df-fun 6501  df-fn 6502  df-f 6503  df-f1 6504  df-fo 6505  df-f1o 6506  df-fv 6507  df-en 8896

This theorem is referenced by:  fndmeng  8983