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Proof step 1 from proof subset-elem-eq-01 (conclusion)

⊦ b ∈ st ∧ st[b] ≠ st1[b] → ¬ st ⊆ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st ⊆ st1, b ∈ st ⊦ st[b] = st1[b]

Rule argument

Additional information