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Proof step 4 from proof subset-put-strong-04 (previous step)

st ⊆ st0 ⊦ ∀ a1. a1 ∈ st[a, d][a0, d0] → a1 ∈ st0[a, d][a0, d0] ∧ 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 5:
st ⊆ st0, a1 ∈ st[a, d][a0, d0], a1 ∈ st0[a, d][a0, d0] ⊦ st[a, d][a0, d0][a1] = st0[a, d][a0, d0][a1]

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;