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Proof step 6 from proof subset-put-strong-04 (previous step)

st ⊆ st0, a1 ∈ st[a, d][a0, d0], a1 ∈ st0[a, d][a0, d0] ⊦ a1 = a0, st[a, d][a0, d0][a1] = st0[a, d][a0, d0][a1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
st ⊆ st0, a1 ∈ st[a, d], a1 ∈ st0[a, d] ⊦ a1 = a0, st[a, d][a1] = st0[a, d][a1]

Rule argument

Additional information

Used simplifier rules: At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);