Metamath Proof Explorer


Theorem rexdif1en

Description: If a set is equinumerous to a nonzero ordinal, then there exists an element in that set such that removing it leaves the set equinumerous to the predecessor of that ordinal. (Contributed by BTernaryTau, 26-Aug-2024) Generalize to all ordinals and avoid ax-un . (Revised by BTernaryTau, 5-Jan-2025)

Ref Expression
Assertion rexdif1en ⊢ M ∈ On ∧ A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M

Proof

Step Hyp Ref Expression
1 encv ⊢ A ≈ suc ⁡ M → A ∈ V ∧ suc ⁡ M ∈ V
2 1 simpld ⊢ A ≈ suc ⁡ M → A ∈ V
3 breng ⊢ A ∈ V ∧ suc ⁡ M ∈ V → A ≈ suc ⁡ M ↔ ∃ f f : A ⟶ 1-1 onto suc ⁡ M
4 1 3 syl ⊢ A ≈ suc ⁡ M → A ≈ suc ⁡ M ↔ ∃ f f : A ⟶ 1-1 onto suc ⁡ M
5 4 ibi ⊢ A ≈ suc ⁡ M → ∃ f f : A ⟶ 1-1 onto suc ⁡ M
6 sucidg ⊢ M ∈ On → M ∈ suc ⁡ M
7 f1ocnvdm ⊢ f : A ⟶ 1-1 onto suc ⁡ M ∧ M ∈ suc ⁡ M → f -1 ⁡ M ∈ A
8 7 ancoms ⊢ M ∈ suc ⁡ M ∧ f : A ⟶ 1-1 onto suc ⁡ M → f -1 ⁡ M ∈ A
9 6 8 sylan ⊢ M ∈ On ∧ f : A ⟶ 1-1 onto suc ⁡ M → f -1 ⁡ M ∈ A
10 9 adantll ⊢ A ∈ V ∧ M ∈ On ∧ f : A ⟶ 1-1 onto suc ⁡ M → f -1 ⁡ M ∈ A
11 vex ⊢ f ∈ V
12 dif1enlem ⊢ f ∈ V ∧ A ∈ V ∧ M ∈ On ∧ f : A ⟶ 1-1 onto suc ⁡ M → A ∖ f -1 ⁡ M ≈ M
13 11 12 mp3anl1 ⊢ A ∈ V ∧ M ∈ On ∧ f : A ⟶ 1-1 onto suc ⁡ M → A ∖ f -1 ⁡ M ≈ M
14 sneq ⊢ x = f -1 ⁡ M → x = f -1 ⁡ M
15 14 difeq2d ⊢ x = f -1 ⁡ M → A ∖ x = A ∖ f -1 ⁡ M
16 15 breq1d ⊢ x = f -1 ⁡ M → A ∖ x ≈ M ↔ A ∖ f -1 ⁡ M ≈ M
17 16 rspcev ⊢ f -1 ⁡ M ∈ A ∧ A ∖ f -1 ⁡ M ≈ M → ∃ x ∈ A A ∖ x ≈ M
18 10 13 17 syl2anc ⊢ A ∈ V ∧ M ∈ On ∧ f : A ⟶ 1-1 onto suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
19 18 ex ⊢ A ∈ V ∧ M ∈ On → f : A ⟶ 1-1 onto suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
20 19 exlimdv ⊢ A ∈ V ∧ M ∈ On → ∃ f f : A ⟶ 1-1 onto suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
21 5 20 syl5 ⊢ A ∈ V ∧ M ∈ On → A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
22 2 21 sylan ⊢ A ≈ suc ⁡ M ∧ M ∈ On → A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
23 22 ancoms ⊢ M ∈ On ∧ A ≈ suc ⁡ M → A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
24 23 syldbl2 ⊢ M ∈ On ∧ A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M