Metamath Proof Explorer


Theorem s4f1o

Description: A length 4 word with mutually different symbols is a one-to-one function onto the set of the symbols. (Contributed by Alexander van der Vekens, 14-Aug-2017)

Ref Expression
Assertion s4f1o ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D → E = ⟨“ ABCD ”⟩ → E : dom ⁡ E ⟶ 1-1 onto A B ∪ C D

Proof

Step Hyp Ref Expression
1 f1oun2prg ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D → 0 A 1 B ∪ 2 C 3 D : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D
2 1 imp ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D → 0 A 1 B ∪ 2 C 3 D : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D
3 2 adantr ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D ∧ E = ⟨“ ABCD ”⟩ → 0 A 1 B ∪ 2 C 3 D : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D
4 s4prop ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → ⟨“ ABCD ”⟩ = 0 A 1 B ∪ 2 C 3 D
5 4 adantr ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D → ⟨“ ABCD ”⟩ = 0 A 1 B ∪ 2 C 3 D
6 5 eqeq2d ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D → E = ⟨“ ABCD ”⟩ ↔ E = 0 A 1 B ∪ 2 C 3 D
7 6 biimpa ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D ∧ E = ⟨“ ABCD ”⟩ → E = 0 A 1 B ∪ 2 C 3 D
8 7 eqcomd ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D ∧ E = ⟨“ ABCD ”⟩ → 0 A 1 B ∪ 2 C 3 D = E
9 8 f1oeq1d ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D ∧ E = ⟨“ ABCD ”⟩ → 0 A 1 B ∪ 2 C 3 D : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D ↔ E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D
10 3 9 mpbid ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D ∧ E = ⟨“ ABCD ”⟩ → E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D
11 dff1o5 ⊢ E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D ↔ E : 0 1 ∪ 2 3 ⟶ 1-1 A B ∪ C D ∧ ran ⁡ E = A B ∪ C D
12 dff12 ⊢ E : 0 1 ∪ 2 3 ⟶ 1-1 A B ∪ C D ↔ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y
13 12 bicomi ⊢ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ↔ E : 0 1 ∪ 2 3 ⟶ 1-1 A B ∪ C D
14 13 anbi1i ⊢ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ∧ ran ⁡ E = A B ∪ C D ↔ E : 0 1 ∪ 2 3 ⟶ 1-1 A B ∪ C D ∧ ran ⁡ E = A B ∪ C D
15 11 14 sylbb2 ⊢ E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D → E : 0 1 ∪ 2 3 ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ∧ ran ⁡ E = A B ∪ C D
16 ffdm ⊢ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D → E : dom ⁡ E ⟶ A B ∪ C D ∧ dom ⁡ E ⊆ 0 1 ∪ 2 3
17 16 simpld ⊢ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D → E : dom ⁡ E ⟶ A B ∪ C D
18 17 anim1i ⊢ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y → E : dom ⁡ E ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y
19 18 anim1i ⊢ E : 0 1 ∪ 2 3 ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ∧ ran ⁡ E = A B ∪ C D → E : dom ⁡ E ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ∧ ran ⁡ E = A B ∪ C D
20 15 19 syl ⊢ E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D → E : dom ⁡ E ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ∧ ran ⁡ E = A B ∪ C D
21 dff12 ⊢ E : dom ⁡ E ⟶ 1-1 A B ∪ C D ↔ E : dom ⁡ E ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y
22 21 anbi1i ⊢ E : dom ⁡ E ⟶ 1-1 A B ∪ C D ∧ ran ⁡ E = A B ∪ C D ↔ E : dom ⁡ E ⟶ A B ∪ C D ∧ ∀ y ∃* x x E y ∧ ran ⁡ E = A B ∪ C D
23 20 22 sylibr ⊢ E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D → E : dom ⁡ E ⟶ 1-1 A B ∪ C D ∧ ran ⁡ E = A B ∪ C D
24 dff1o5 ⊢ E : dom ⁡ E ⟶ 1-1 onto A B ∪ C D ↔ E : dom ⁡ E ⟶ 1-1 A B ∪ C D ∧ ran ⁡ E = A B ∪ C D
25 23 24 sylibr ⊢ E : 0 1 ∪ 2 3 ⟶ 1-1 onto A B ∪ C D → E : dom ⁡ E ⟶ 1-1 onto A B ∪ C D
26 10 25 syl ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S ∧ A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D ∧ E = ⟨“ ABCD ”⟩ → E : dom ⁡ E ⟶ 1-1 onto A B ∪ C D
27 26 exp31 ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ∧ D ∈ S → A ≠ B ∧ A ≠ C ∧ A ≠ D ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D → E = ⟨“ ABCD ”⟩ → E : dom ⁡ E ⟶ 1-1 onto A B ∪ C D