Theorem fundmeng 9013

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 6541	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 5879	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 22	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 5113	. . . 4 ⊢ (𝑥 = 𝐹 → (dom 𝑥 ≈ 𝑥 ↔ dom 𝐹 ≈ 𝐹))
5	1, 4	imbi12d 346	. . 3 ⊢ (𝑥 = 𝐹 → ((Fun 𝑥 → dom 𝑥 ≈ 𝑥) ↔ (Fun 𝐹 → dom 𝐹 ≈ 𝐹)))
6		vex 3458	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 9012	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 3522	. 2 ⊢ (𝐹 ∈ 𝑉 → (Fun 𝐹 → dom 𝐹 ≈ 𝐹))
9	8	imp 410	1 ⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)

Syntax hints:   → wi 4   ∧ wa 399   = wceq 1560   ∈ wcel 2142   class class class wbr 5100  dom cdm 5647  Fun wfun 6515   ≈ cen 8924

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734  ax-sep 5246  ax-nul 5256  ax-pow 5322  ax-pr 5390  ax-un 7718

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-nf 1804  df-sb 2091  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3077  df-rex 3087  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4481  df-pw 4557  df-sn 4583  df-pr 4585  df-op 4589  df-uni 4866  df-int 4906  df-br 5101  df-opab 5163  df-mpt 5182  df-id 5542  df-xp 5653  df-rel 5654  df-cnv 5655  df-co 5656  df-dm 5657  df-rn 5658  df-iota 6477  df-fun 6523  df-fn 6524  df-f 6525  df-f1 6526  df-fo 6527  df-f1o 6528  df-fv 6529  df-en 8928

This theorem is referenced by:  fndmeng  9016