Metamath Proof Explorer


Theorem r1sssucd

Description: Deductive form of r1sssuc . (Contributed by Noam Pasman, 19-Jan-2025)

Ref Expression
Hypothesis r1sssucd.1 ⊢ φ → A ∈ On
Assertion r1sssucd ⊢ φ → R1 ⁡ A ⊆ R1 ⁡ suc ⁡ A

Proof

Step Hyp Ref Expression
1 r1sssucd.1 ⊢ φ → A ∈ On
2 r1sssuc ⊢ A ∈ On → R1 ⁡ A ⊆ R1 ⁡ suc ⁡ A
3 1 2 syl ⊢ φ → R1 ⁡ A ⊆ R1 ⁡ suc ⁡ A