Theorem funss 5237

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 4715	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (Rel 𝐵 → Rel 𝐴))
2		coss1 4784	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐴))
3		cnvss 4802	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → ◡𝐴 ⊆ ◡𝐵)
4		coss2 4785	. . . . . 6 ⊢ (◡𝐴 ⊆ ◡𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
5	3, 4	syl 14	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝐵 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
6	2, 5	sstrd 3167	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∘ ◡𝐴) ⊆ (𝐵 ∘ ◡𝐵))
7		sstr2 3164	. . . 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 5220	. 2 ⊢ (Fun 𝐵 ↔ (Rel 𝐵 ∧ (𝐵 ∘ ◡𝐵) ⊆ I ))
11		df-fun 5220	. 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 3131   I cid 4290  ^◡ccnv 4627   ∘ ccom 4632  Rel wrel 4633  Fun wfun 5212

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-in 3137  df-ss 3144  df-br 4006  df-opab 4067  df-rel 4635  df-cnv 4636  df-co 4637  df-fun 5220

This theorem is referenced by:  funeq  5238  funopab4  5255  funres  5259  fun0  5276  funcnvcnv  5277  funin  5289  funres11  5290  foimacnv  5481  tfrlemibfn  6331  tfr1onlembfn  6347  tfrcllembfn  6360  strslssd  12511  strle1g  12567