Theorem funss 4863

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 4370	. . 3 ⊢ (A ⊆ B → (Rel B → Rel A))
2		coss1 4434	. . . . 5 ⊢ (A ⊆ B → (A ∘ ◡A) ⊆ (B ∘ ◡A))
3		cnvss 4451	. . . . . 6 ⊢ (A ⊆ B → ◡A ⊆ ◡B)
4		coss2 4435	. . . . . 6 ⊢ (◡A ⊆ ◡B → (B ∘ ◡A) ⊆ (B ∘ ◡B))
5	3, 4	syl 14	. . . . 5 ⊢ (A ⊆ B → (B ∘ ◡A) ⊆ (B ∘ ◡B))
6	2, 5	sstrd 2949	. . . 4 ⊢ (A ⊆ B → (A ∘ ◡A) ⊆ (B ∘ ◡B))
7		sstr2 2946	. . . 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 4847	. 2 ⊢ (Fun B ↔ (Rel B ∧ (B ∘ ◡B) ⊆ I ))
11		df-fun 4847	. 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 2911   I cid 4016  ^◡ccnv 4287   ∘ ccom 4292  Rel wrel 4293  Fun wfun 4839

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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-in 2918  df-ss 2925  df-br 3756  df-opab 3810  df-rel 4295  df-cnv 4296  df-co 4297  df-fun 4847

This theorem is referenced by:  funeq  4864  funopab4  4880  funres  4884  fun0  4900  funcnvcnv  4901  funin  4913  funres11  4914  foimacnv  5087  tfrlemibfn  5883