Theorem funss 5370

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 4836	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Rel 𝐵 → Rel 𝐴))
2		coss1 4909	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐴))
3		cnvss 4927	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
4		coss2 4910	. . . . . 6 ⊢ (◡𝐴 ⊆ ◡𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
5	3, 4	syl 14	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
6	2, 5	sstrd 3247	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
7		sstr2 3244	. . . 4 ⊢ ((𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵) → ((𝐵 ∘ ◡𝐵) ⊆ I → (𝐴 ∘ ◡𝐴) ⊆ I ))
8	6, 7	syl 14	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝐵 ∘ ◡𝐵) ⊆ I → (𝐴 ∘ ◡𝐴) ⊆ I ))
9	1, 8	anim12d 335	. 2 ⊢ (𝐴 ⊆ 𝐵 → ((Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ) → (Rel 𝐴 ∧ (𝐴 ∘ ◡𝐴) ⊆ I )))
10		df-fun 5353	. 2 ⊢ (Fun 𝐵 ↔ (Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ))
11		df-fun 5353	. 2 ⊢ (Fun 𝐴 ↔ (Rel 𝐴 ∧ (𝐴 ∘ ◡𝐴) ⊆ I ))
12	9, 10, 11	3imtr4g 205	1 ⊢ (𝐴 ⊆ 𝐵 → (Fun 𝐵 → Fun 𝐴))

Syntax hints:   → wi 4   ∧ wa 104   ⊆ wss 3210   I cid 4408  ^◡ccnv 4747   ∘ ccom 4752  Rel wrel 4753  Fun wfun 5345

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-in 3216  df-ss 3223  df-br 4109  df-opab 4171  df-rel 4755  df-cnv 4756  df-co 4757  df-fun 5353

This theorem is referenced by:  funeq  5371  funopab4  5388  funres  5392  fun0  5413  funcnvcnv  5414  funin  5426  funres11  5427  foimacnv  5631  funsssuppss  6457  tfrlemibfn  6558  tfr1onlembfn  6574  tfrcllembfn  6587  strslssd  13251  strle1g  13311  subgrfun  16254