Theorem fundmeng 9025

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 6556	. . . 4 ⊢ (𝑥 = 𝐹 → (Fun 𝑥 ↔ Fun 𝐹))
2		dmeq 5893	. . . . 5 ⊢ (𝑥 = 𝐹 → dom 𝑥 = dom 𝐹)
3		id 23	. . . . 5 ⊢ (𝑥 = 𝐹 → 𝑥 = 𝐹)
4	2, 3	breq12d 5122	. . . 4 ⊢ (𝑥 = 𝐹 → (dom 𝑥 ≈ 𝑥 ↔ dom 𝐹 ≈ 𝐹))
5	1, 4	imbi12d 347	. . 3 ⊢ (𝑥 = 𝐹 → ((Fun 𝑥 → dom 𝑥 ≈ 𝑥) ↔ (Fun 𝐹 → dom 𝐹 ≈ 𝐹)))
6		vex 3459	. . . 4 ⊢ 𝑥 ∈ V
7	6	fundmen 9024	. . 3 ⊢ (Fun 𝑥 → dom 𝑥 ≈ 𝑥)
8	5, 7	vtoclg 3522	. 2 ⊢ (𝐹 ∈ 𝑉 → (Fun 𝐹 → dom 𝐹 ≈ 𝐹))
9	8	imp 411	1 ⊢ ((𝐹 ∈ 𝑉 ∧ Fun 𝐹) → dom 𝐹 ≈ 𝐹)

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1570   ∈ wcel 2143   class class class wbr 5109  dom cdm 5661  Fun wfun 6530   ≈ cen 8936

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735  ax-sep 5257  ax-nul 5269  ax-pow 5336  ax-pr 5404  ax-un 7732

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-nf 1814  df-sb 2097  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-ne 2959  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-pw 4564  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4873  df-int 4913  df-br 5110  df-opab 5174  df-mpt 5193  df-id 5556  df-xp 5667  df-rel 5668  df-cnv 5669  df-co 5670  df-dm 5671  df-rn 5672  df-iota 6492  df-fun 6538  df-fn 6539  df-f 6540  df-f1 6541  df-fo 6542  df-f1o 6543  df-fv 6544  df-en 8940

This theorem is referenced by:  fndmeng  9028