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Proof step 1 from proof subset-elem-eq (conclusion)

⊦ b ∈ st ∧ st ⊆ st1 → st[b] = st1[b]

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: subset-elem-eq-02: ⊦ b ∈ st ∧ st ⊆ st1 → (st[b] ≠ st1[b] ↔ false);