Theorem funss 5273

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 4746	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Rel 𝐵 → Rel 𝐴))
2		coss1 4817	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐴))
3		cnvss 4835	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
4		coss2 4818	. . . . . 6 ⊢ (◡𝐴 ⊆ ◡𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
5	3, 4	syl 14	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
6	2, 5	sstrd 3189	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
7		sstr2 3186	. . . 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 5256	. 2 ⊢ (Fun 𝐵 ↔ (Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ))
11		df-fun 5256	. 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 3153   I cid 4319  ^◡ccnv 4658   ∘ ccom 4663  Rel wrel 4664  Fun wfun 5248

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-in 3159  df-ss 3166  df-br 4030  df-opab 4091  df-rel 4666  df-cnv 4667  df-co 4668  df-fun 5256

This theorem is referenced by:  funeq  5274  funopab4  5291  funres  5295  fun0  5312  funcnvcnv  5313  funin  5325  funres11  5326  foimacnv  5518  tfrlemibfn  6381  tfr1onlembfn  6397  tfrcllembfn  6410  strslssd  12665  strle1g  12724