Theorem fundmeng 6958

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 5337	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 4922	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 19	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 4095	. . . 4 ⊢ (𝑥 = 𝐹 → (dom 𝑥 ≈ 𝑥 ↔ dom 𝐹 ≈ 𝐹))
5	1, 4	imbi12d 234	. . 3 ⊢ (𝑥 = 𝐹 → ((Fun 𝑥 → dom 𝑥 ≈ 𝑥) ↔ (Fun 𝐹 → dom 𝐹 ≈ 𝐹)))
6		vex 2802	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 6957	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 2861	. 2 ⊢ (𝐹 ∈ 𝑉 → (Fun 𝐹 → dom 𝐹 ≈ 𝐹))
9	8	imp 124	1 ⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1395   ∈ wcel 2200   class class class wbr 4082  dom cdm 4718  Fun wfun 5311   ≈ cen 6883

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-13 2202  ax-14 2203  ax-ext 2211  ax-sep 4201  ax-pow 4257  ax-pr 4292  ax-un 4523

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-eu 2080  df-mo 2081  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2801  df-sbc 3029  df-un 3201  df-in 3203  df-ss 3210  df-pw 3651  df-sn 3672  df-pr 3673  df-op 3675  df-uni 3888  df-int 3923  df-br 4083  df-opab 4145  df-mpt 4146  df-id 4383  df-xp 4724  df-rel 4725  df-cnv 4726  df-co 4727  df-dm 4728  df-rn 4729  df-iota 5277  df-fun 5319  df-fn 5320  df-f 5321  df-f1 5322  df-fo 5323  df-f1o 5324  df-fv 5325  df-en 6886

This theorem is referenced by:  fndmeng  6961  fundmfi  7100  usgrsizedgen  16005