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Proof step 10 from proof subset-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a ∈ st0, a0 ∈ st, ∀ a1. a1 ∈ st -- a → a1 ∈ st0 ∧ st[a1] = st0[a1] ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

Rule argument

Additional information

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