Theorem funss 6597

Description: Subclass theorem for function predicate. (Contributed by NM, 16-Aug-1994.) (Proof shortened by Mario Carneiro, 24-Jun-2014.)

Assertion

Ref	Expression
funss	⊢ (𝐴 ⊆ 𝐵 → (Fun 𝐵 → Fun 𝐴))

Proof of Theorem funss

Step	Hyp	Ref	Expression
1		relss 5805	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Rel 𝐵 → Rel 𝐴))
2		coss1 5880	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐴))
3		cnvss 5897	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
4		coss2 5881	. . . . . 6 ⊢ (◡𝐴 ⊆ ◡𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
5	3, 4	syl 17	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
6	2, 5	sstrd 4019	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
7		sstr2 4015	. . . 4 ⊢ ((𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵) → ((𝐵 ∘ ◡𝐵) ⊆ I → (𝐴 ∘ ◡𝐴) ⊆ I ))
8	6, 7	syl 17	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝐵 ∘ ◡𝐵) ⊆ I → (𝐴 ∘ ◡𝐴) ⊆ I ))
9	1, 8	anim12d 608	. 2 ⊢ (𝐴 ⊆ 𝐵 → ((Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ) → (Rel 𝐴 ∧ (𝐴 ∘ ◡𝐴) ⊆ I )))
10		df-fun 6575	. 2 ⊢ (Fun 𝐵 ↔ (Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ))
11		df-fun 6575	. 2 ⊢ (Fun 𝐴 ↔ (Rel 𝐴 ∧ (𝐴 ∘ ◡𝐴) ⊆ I ))
12	9, 10, 11	3imtr4g 296	1 ⊢ (𝐴 ⊆ 𝐵 → (Fun 𝐵 → Fun 𝐴))

Syntax hints:   → wi 4   ∧ wa 395   ⊆ wss 3976   I cid 5592  ^◡ccnv 5699   ∘ ccom 5704  Rel wrel 5705  Fun wfun 6567

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-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ss 3993  df-br 5167  df-opab 5229  df-rel 5707  df-cnv 5708  df-co 5709  df-fun 6575

This theorem is referenced by:  funeq  6598  funopab4  6615  funres  6620  fun0  6643  funcnvcnv  6645  funin  6654  funres11  6655  foimacnv  6879  funelss  8088  funsssuppss  8231  fsuppss  9452  strle1  17205  strssd  17253  pjpm  21751  subgrfun  29316  setrecsss  48793