Metamath Proof Explorer


Theorem suceqd

Description: Deduction associated with suceq . (Contributed by Rohan Ridenour, 8-Aug-2023)

Ref Expression
Hypothesis suceqd.1 ⊢ φ → A = B
Assertion suceqd ⊢ φ → suc ⁡ A = suc ⁡ B

Proof

Step Hyp Ref Expression
1 suceqd.1 ⊢ φ → A = B
2 1 sneqd ⊢ φ → A = B
3 1 2 uneq12d ⊢ φ → A ∪ A = B ∪ B
4 df-suc ⊢ suc ⁡ A = A ∪ A
5 df-suc ⊢ suc ⁡ B = B ∪ B
6 3 4 5 3eqtr4g ⊢ φ → suc ⁡ A = suc ⁡ B