Metamath Proof Explorer


Theorem mopre

Description: There is at most one predecessor of N . (Contributed by Peter Mazsa, 12-Jan-2026)

Ref Expression
Assertion mopre ⊢ ∃* m suc ⁡ m = N

Proof

Step Hyp Ref Expression
1 eqtr3 ⊢ suc ⁡ m = N ∧ suc ⁡ l = N → suc ⁡ m = suc ⁡ l
2 suc11reg ⊢ suc ⁡ m = suc ⁡ l ↔ m = l
3 1 2 sylib ⊢ suc ⁡ m = N ∧ suc ⁡ l = N → m = l
4 3 gen2 ⊢ ∀ m ∀ l suc ⁡ m = N ∧ suc ⁡ l = N → m = l
5 suceq ⊢ m = l → suc ⁡ m = suc ⁡ l
6 5 eqeq1d ⊢ m = l → suc ⁡ m = N ↔ suc ⁡ l = N
7 6 mo4 ⊢ ∃* m suc ⁡ m = N ↔ ∀ m ∀ l suc ⁡ m = N ∧ suc ⁡ l = N → m = l
8 4 7 mpbir ⊢ ∃* m suc ⁡ m = N