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Proof step 3 from proof empty-oplus (previous step)

⊦ ∀ a. (a ∈ ∅ ⊕ st ↔ a ∈ st) ∧ (a ∈ ∅ ⊕ st → (∅ ⊕ st)[a] = st[a])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (a ∈ st → a ∈ ∅ ⊕ st) ∧ (a ∈ ∅ ⊕ st → (∅ ⊕ st)[a] = st[a])

Rule argument

Additional information

Used simplifier rules: oplus-l-in: ⊦ a ∈ st0 ⊕ st1 ↔ a ∈ st0 ∨ a ∈ st1;%In-empty: ⊦ ¬ a ∈ ∅; from specification store