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Proof step 2 from proof sub (previous step)

a ∈ s, s ⊆ st.keys ⊦ a ∈ st

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Keys: ⊦ a ∈ st.keys ↔ a ∈ st;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic