Metamath Proof Explorer


Theorem setind

Description: Set (epsilon) induction. Theorem 5.22 of TakeutiZaring p. 21. (Contributed by NM, 17-Sep-2003)

Ref Expression
Assertion setind ⊢ ∀ x x ⊆ A → x ∈ A → A = V

Proof

Step Hyp Ref Expression
1 ssindif0 ⊢ y ⊆ A ↔ y ∩ V ∖ A = ∅
2 sseq1 ⊢ x = y → x ⊆ A ↔ y ⊆ A
3 eleq1w ⊢ x = y → x ∈ A ↔ y ∈ A
4 2 3 imbi12d ⊢ x = y → x ⊆ A → x ∈ A ↔ y ⊆ A → y ∈ A
5 4 spvv ⊢ ∀ x x ⊆ A → x ∈ A → y ⊆ A → y ∈ A
6 1 5 biimtrrid ⊢ ∀ x x ⊆ A → x ∈ A → y ∩ V ∖ A = ∅ → y ∈ A
7 eldifn ⊢ y ∈ V ∖ A → ¬ y ∈ A
8 6 7 nsyli ⊢ ∀ x x ⊆ A → x ∈ A → y ∈ V ∖ A → ¬ y ∩ V ∖ A = ∅
9 8 imp ⊢ ∀ x x ⊆ A → x ∈ A ∧ y ∈ V ∖ A → ¬ y ∩ V ∖ A = ∅
10 9 nrexdv ⊢ ∀ x x ⊆ A → x ∈ A → ¬ ∃ y ∈ V ∖ A y ∩ V ∖ A = ∅
11 zfregs ⊢ V ∖ A ≠ ∅ → ∃ y ∈ V ∖ A y ∩ V ∖ A = ∅
12 11 necon1bi ⊢ ¬ ∃ y ∈ V ∖ A y ∩ V ∖ A = ∅ → V ∖ A = ∅
13 10 12 syl ⊢ ∀ x x ⊆ A → x ∈ A → V ∖ A = ∅
14 vdif0 ⊢ A = V ↔ V ∖ A = ∅
15 13 14 sylibr ⊢ ∀ x x ⊆ A → x ∈ A → A = V