Theorem fundmeng 8198

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 6069	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 5479	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 22	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 4817	. . . 4 ⊢ (𝑥 = 𝐹 → (dom 𝑥 ≈ 𝑥 ↔ dom 𝐹 ≈ 𝐹))
5	1, 4	imbi12d 333	. . 3 ⊢ (𝑥 = 𝐹 → ((Fun 𝑥 → dom 𝑥 ≈ 𝑥) ↔ (Fun 𝐹 → dom 𝐹 ≈ 𝐹)))
6		vex 3343	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 8197	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 3406	. 2 ⊢ (𝐹 ∈ 𝑉 → (Fun 𝐹 → dom 𝐹 ≈ 𝐹))
9	8	imp 444	1 ⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)

Syntax hints:   → wi 4   ∧ wa 383   = wceq 1632   ∈ wcel 2139   class class class wbr 4804  dom cdm 5266  Fun wfun 6043   ≈ cen 8120

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-8 2141  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740  ax-sep 4933  ax-nul 4941  ax-pow 4992  ax-pr 5055  ax-un 7115

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1635  df-ex 1854  df-nf 1859  df-sb 2047  df-eu 2611  df-mo 2612  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-ral 3055  df-rex 3056  df-rab 3059  df-v 3342  df-sbc 3577  df-dif 3718  df-un 3720  df-in 3722  df-ss 3729  df-nul 4059  df-if 4231  df-pw 4304  df-sn 4322  df-pr 4324  df-op 4328  df-uni 4589  df-int 4628  df-br 4805  df-opab 4865  df-mpt 4882  df-id 5174  df-xp 5272  df-rel 5273  df-cnv 5274  df-co 5275  df-dm 5276  df-rn 5277  df-iota 6012  df-fun 6051  df-fn 6052  df-f 6053  df-f1 6054  df-fo 6055  df-f1o 6056  df-fv 6057  df-en 8124

This theorem is referenced by:  fndmeng  8201