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Proof step 6 from proof subset-delete-01 (previous step)

∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a] ⊦ ∀ a0. a0 ∈ st -- a → a0 ∈ st0 ∧ (st -- a)[a0] = st0[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 ∈ st, ∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a] ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

Rule argument

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a];