Theorem ffdm 6699

Description: A mapping is a partial function. (Contributed by NM, 25-Nov-2007.)

Assertion

Ref	Expression
ffdm	⊢ (𝐹:𝐴⟶𝐵 → (𝐹:dom 𝐹⟶𝐵 ∧ dom 𝐹 ⊆ 𝐴))

Proof of Theorem ffdm

Step	Hyp	Ref	Expression
1		fdm 6679	. . . 4 ⊢ (𝐹:𝐴⟶𝐵 → dom 𝐹 = 𝐴)
2	1	feq2d 6654	. . 3 ⊢ (𝐹:𝐴⟶𝐵 → (𝐹:dom 𝐹⟶𝐵 ↔ 𝐹:𝐴⟶𝐵))
3	2	ibir 268	. 2 ⊢ (𝐹:𝐴⟶𝐵 → 𝐹:dom 𝐹⟶𝐵)
4		eqimss 3994	. . 3 ⊢ (dom 𝐹 = 𝐴 → dom 𝐹 ⊆ 𝐴)
5	1, 4	syl 17	. 2 ⊢ (𝐹:𝐴⟶𝐵 → dom 𝐹 ⊆ 𝐴)
6	3, 5	jca 511	1 ⊢ (𝐹:𝐴⟶𝐵 → (𝐹:dom 𝐹⟶𝐵 ∧ dom 𝐹 ⊆ 𝐴))

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1542   ⊆ wss 3903  dom cdm 5632  ⟶wf 6496

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-cleq 2729  df-ss 3920  df-fn 6503  df-f 6504

This theorem is referenced by:  ffdmd  6700  smoiso  8304  s4f1o  14853  islindf2  21781  fourierdlem92  46556  fouriersw  46589  etransclem2  46594