Theorem funss 4834

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 4342	. . 3 ⊢ (A ⊆ B → (Rel B → Rel A))
2		coss1 4406	. . . . 5 ⊢ (A ⊆ B → (A ∘ ◡A) ⊆ (B ∘ ◡A))
3		cnvss 4423	. . . . . 6 ⊢ (A ⊆ B → ◡A ⊆ ◡B)
4		coss2 4407	. . . . . 6 ⊢ (◡A ⊆ ◡B → (B ∘ ◡A) ⊆ (B ∘ ◡B))
5	3, 4	syl 14	. . . . 5 ⊢ (A ⊆ B → (B ∘ ◡A) ⊆ (B ∘ ◡B))
6	2, 5	sstrd 2923	. . . 4 ⊢ (A ⊆ B → (A ∘ ◡A) ⊆ (B ∘ ◡B))
7		sstr2 2920	. . . 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 318	. 2 ⊢ (A ⊆ B → ((Rel B ∧ (B ∘ ◡B) ⊆ I ) → (Rel A ∧ (A ∘ ◡A) ⊆ I )))
10		df-fun 4819	. 2 ⊢ (Fun B ↔ (Rel B ∧ (B ∘ ◡B) ⊆ I ))
11		df-fun 4819	. 2 ⊢ (Fun A ↔ (Rel A ∧ (A ∘ ◡A) ⊆ I ))
12	9, 10, 11	3imtr4g 194	1 ⊢ (A ⊆ B → (Fun B → Fun A))

Syntax hints:   → wi 4   ∧ wa 97   ⊆ wss 2885   I cid 3988  ^◡ccnv 4259   ∘ ccom 4264  Rel wrel 4265  Fun wfun 4811

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 614  ax-5 1309  ax-7 1310  ax-gen 1311  ax-ie1 1355  ax-ie2 1356  ax-8 1368  ax-10 1369  ax-11 1370  ax-i12 1371  ax-bnd 1372  ax-4 1373  ax-17 1392  ax-i9 1396  ax-ial 1400  ax-i5r 1401  ax-ext 1995

This theorem depends on definitions:  df-bi 110  df-nf 1323  df-sb 1619  df-clab 2000  df-cleq 2006  df-clel 2009  df-nfc 2140  df-in 2892  df-ss 2899  df-br 3728  df-opab 3782  df-rel 4267  df-cnv 4268  df-co 4269  df-fun 4819

This theorem is referenced by:  funeq  4835  funopab4  4851  funres  4855  fun0  4871  funcnvcnv  4872  funin  4884  funres11  4885  foimacnv  5057  tfrlemibfn  5851