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Proof step 2 from proof subset-elem-eq-01 (
previous step
)
b ∈ st, st ⊆ st1 ⊦ st[b] = st1[b]
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
b ∈ st, st ⊆ st1 ⊦ st[b] = st1[b]
Rule argument
Additional information