Theorem funss 6544

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 5759	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Rel 𝐵 → Rel 𝐴))
2		coss1 5832	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐴))
3		cnvss 5849	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
4		coss2 5833	. . . . . 6 ⊢ (◡𝐴 ⊆ ◡𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
5	3, 4	syl 18	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
6	2, 5	sstrd 3949	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
7		sstr2 3946	. . . 4 ⊢ ((𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵) → ((𝐵 ∘ ◡𝐵) ⊆ I → (𝐴 ∘ ◡𝐴) ⊆ I ))
8	6, 7	syl 18	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝐵 ∘ ◡𝐵) ⊆ I → (𝐴 ∘ ◡𝐴) ⊆ I ))
9	1, 8	anim12d 620	. 2 ⊢ (𝐴 ⊆ 𝐵 → ((Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ) → (Rel 𝐴 ∧ (𝐴 ∘ ◡𝐴) ⊆ I )))
10		df-fun 6527	. 2 ⊢ (Fun 𝐵 ↔ (Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ))
11		df-fun 6527	. 2 ⊢ (Fun 𝐴 ↔ (Rel 𝐴 ∧ (𝐴 ∘ ◡𝐴) ⊆ I ))
12	9, 10, 11	3imtr4g 299	1 ⊢ (𝐴 ⊆ 𝐵 → (Fun 𝐵 → Fun 𝐴))

Syntax hints:   → wi 4   ∧ wa 400   ⊆ wss 3907   I cid 5546  ^◡ccnv 5651   ∘ ccom 5656  Rel wrel 5657  Fun wfun 6519

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1803  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-ss 3924  df-br 5106  df-opab 5168  df-rel 5659  df-cnv 5660  df-co 5661  df-fun 6527

This theorem is referenced by:  funeq  6545  funopab4  6562  funres  6567  fun0  6590  funcnvcnv  6592  funin  6601  funres11  6602  foimacnv  6828  funelss  8032  funsssuppss  8174  fsuppss  9331  strle1  17208  strssd  17255  pjpm  21818  subgrfun  29540  setrecsss  50330