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Proof step 4 from proof subset-put-strong-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
st ⊆ st0, a2 ∈ st[a, d][a0, d0][a1, d1], a2 ∈ st0[a, d][a0, d0][a1, d1] ⊦ st[a, d][a0, d0][a1, d1][a2] = st0[a, d][a0, d0][a1, d1][a2]

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%superset-elem-in-01: ⊦ b ∈ st ∧ ¬ b ∈ st0 → ¬ st ⊆ st0;