Metamath Proof Explorer


Theorem efmnd2hash

Description: The monoid of endofunctions on a (proper) pair has cardinality 4 . (Contributed by AV, 18-Feb-2024)

Ref Expression
Hypotheses efmnd1bas.1 ⊢ G = EndoFMnd ⁡ A
efmnd1bas.2 ⊢ B = Base G
efmnd2bas.0 ⊢ A = I J
Assertion efmnd2hash ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → B = 4

Proof

Step Hyp Ref Expression
1 efmnd1bas.1 ⊢ G = EndoFMnd ⁡ A
2 efmnd1bas.2 ⊢ B = Base G
3 efmnd2bas.0 ⊢ A = I J
4 prfi ⊢ I J ∈ Fin
5 3 4 eqeltri ⊢ A ∈ Fin
6 1 2 efmndhash ⊢ A ∈ Fin → B = A A
7 5 6 ax-mp ⊢ B = A A
8 3 fveq2i ⊢ A = I J
9 elex ⊢ I ∈ V → I ∈ V
10 elex ⊢ J ∈ W → J ∈ V
11 id ⊢ I ≠ J → I ≠ J
12 9 10 11 3anim123i ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → I ∈ V ∧ J ∈ V ∧ I ≠ J
13 hashprb ⊢ I ∈ V ∧ J ∈ V ∧ I ≠ J ↔ I J = 2
14 12 13 sylib ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → I J = 2
15 8 14 eqtrid ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → A = 2
16 15 15 oveq12d ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → A A = 2 2
17 sq2 ⊢ 2 2 = 4
18 16 17 eqtrdi ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → A A = 4
19 7 18 eqtrid ⊢ I ∈ V ∧ J ∈ W ∧ I ≠ J → B = 4