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Proof step 5 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 6:
a1 ∈ st[a, d][a0, d0], a1 ∈ st0[a, d][a0, d0], st ⊆ st0 ⊦ 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;%superset-elem-in-01: ⊦ b ∈ st ∧ ¬ b ∈ st0 → ¬ st ⊆ st0;