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Proof step 5 from proof subset-put-strong-03 (previous step)

st ⊆ st0 ⊦ ∀ a0. a0 ∈ st[a, d] → a0 ∈ st0[a, d] ∧ st[a, d][a0] = st0[a, d][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ st[a, d], st ⊆ st0 ⊦ a0 ∈ st0[a, d] ∧ st[a, d][a0] = st0[a, d][a0]

Rule argument

Additional information