Theorem funss 2682

Description: Subclass theorem for function predicate.

Assertion

Ref	Expression
funss	⊢ (A ⊆ B → (Fun B → Fun A))

Proof of Theorem funss

Step	Hyp	Ref	Expression
1		ssrel 2479	. . . 4 ⊢ (A ⊆ B → (Rel B → Rel A))
2		funrel 2681	. . . 4 ⊢ (Fun B → Rel B)
3	1, 2	syl5 22	. . 3 ⊢ (A ⊆ B → (Fun B → Rel A))
4		ssel 1502	. . . . . . . 8 ⊢ (A ⊆ B → (⟨x, y⟩ ∈ A → ⟨x, y⟩ ∈ B))
5	4	syl4d 28	. . . . . . 7 ⊢ (A ⊆ B → ((⟨x, y⟩ ∈ B → y = z) → (⟨x, y⟩ ∈ A → y = z)))
6	5	19.20dv 946	. . . . . 6 ⊢ (A ⊆ B → (∀y(⟨x, y⟩ ∈ B → y = z) → ∀y(⟨x, y⟩ ∈ A → y = z)))
7	6	19.22dv 947	. . . . 5 ⊢ (A ⊆ B → (∃z∀y(⟨x, y⟩ ∈ B → y = z) → ∃z∀y(⟨x, y⟩ ∈ A → y = z)))
8	7	19.20dv 946	. . . 4 ⊢ (A ⊆ B → (∀x∃z∀y(⟨x, y⟩ ∈ B → y = z) → ∀x∃z∀y(⟨x, y⟩ ∈ A → y = z)))
9		dffun5 2677	. . . . 5 ⊢ (Fun B ↔ (Rel B ∧ ∀x∃z∀y(⟨x, y⟩ ∈ B → y = z)))
10	9	pm3.27bd 263	. . . 4 ⊢ (Fun B → ∀x∃z∀y(⟨x, y⟩ ∈ B → y = z))
11	8, 10	syl5 22	. . 3 ⊢ (A ⊆ B → (Fun B → ∀x∃z∀y(⟨x, y⟩ ∈ A → y = z)))
12	3, 11	jcad 455	. 2 ⊢ (A ⊆ B → (Fun B → (Rel A ∧ ∀x∃z∀y(⟨x, y⟩ ∈ A → y = z))))
13		dffun5 2677	. 2 ⊢ (Fun A ↔ (Rel A ∧ ∀x∃z∀y(⟨x, y⟩ ∈ A → y = z)))
14	12, 13	syl6ibr 186	1 ⊢ (A ⊆ B → (Fun B → Fun A))

Syntax hints:   → wi 2   ∧ wa 196  ∀wal 672  ∃wex 678   = weq 797   ∈ wcel 1092   ⊆ wss 1487  ⟨cop 1810  Rel wrel 2415  Fun wfun 2416

This theorem is referenced by:  funeq 2683  fun0 2691  funres 2697  funcnvcnv 2701  funres11 2709  fodom 3613

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-eu 1009  df-mo 1010  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-opab 2098  df-id 2125  df-rel 2425  df-cnv 2426  df-co 2427  df-fun 2432

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