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Proof step 7 from proof subset-put-strong-05 (previous step)

a2 ∈ st0[a, d][a0, d0][a1, d1], a2 ∈ st[a, d][a0, d0][a1, d1], st ⊆ st0 ⊦ st[a, d][a0, d0][a1, d1][a2] = st0[a, d][a0, d0][a1, d1][a2], a2 = a1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a2 ∈ st[a, d][a0, d0], a2 ∈ st0[a, d][a0, d0], st ⊆ st0 ⊦ st[a, d][a0, d0][a2] = st0[a, d][a0, d0][a2], a2 = a1

Rule argument

Additional information

Used simplifier rules: At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;