Metamath Proof Explorer


Theorem wsuccl

Description: If X is a set with an R successor in A , then its well-founded successor is a member of A . (Contributed by Scott Fenton, 15-Jun-2018) (Proof shortened by AV, 10-Oct-2021)

Ref Expression
Hypotheses wsuccl.1 ⊢ φ → R We A
wsuccl.2 ⊢ φ → R Se A
wsuccl.3 ⊢ φ → X ∈ V
wsuccl.4 ⊢ φ → ∃ y ∈ A X R y
Assertion wsuccl ⊢ φ → wsuc R A X ∈ A

Proof

Step Hyp Ref Expression
1 wsuccl.1 ⊢ φ → R We A
2 wsuccl.2 ⊢ φ → R Se A
3 wsuccl.3 ⊢ φ → X ∈ V
4 wsuccl.4 ⊢ φ → ∃ y ∈ A X R y
5 df-wsuc ⊢ wsuc R A X = inf Pred R -1 A X A R
6 weso ⊢ R We A → R Or A
7 1 6 syl ⊢ φ → R Or A
8 1 2 3 4 wsuclem ⊢ φ → ∃ a ∈ A ∀ b ∈ Pred R -1 A X ¬ b R a ∧ ∀ b ∈ A a R b → ∃ c ∈ Pred R -1 A X c R b
9 7 8 infcl ⊢ φ → inf Pred R -1 A X A R ∈ A
10 5 9 eqeltrid ⊢ φ → wsuc R A X ∈ A