Theorem fundmeng 9097

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 6598	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 5928	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 22	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 5179	. . . 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 3492	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 9096	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 3566	. 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 1537   ∈ wcel 2108   class class class wbr 5166  dom cdm 5700  Fun wfun 6567   ≈ cen 9000

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pow 5383  ax-pr 5447  ax-un 7770

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-mo 2543  df-eu 2572  df-clab 2718  df-cleq 2732  df-clel 2819  df-nfc 2895  df-ne 2947  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-if 4549  df-pw 4624  df-sn 4649  df-pr 4651  df-op 4655  df-uni 4932  df-int 4971  df-br 5167  df-opab 5229  df-mpt 5250  df-id 5593  df-xp 5706  df-rel 5707  df-cnv 5708  df-co 5709  df-dm 5710  df-rn 5711  df-iota 6525  df-fun 6575  df-fn 6576  df-f 6577  df-f1 6578  df-fo 6579  df-f1o 6580  df-fv 6581  df-en 9004

This theorem is referenced by:  fndmeng  9100