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Proof step 10 from proof subset-put-01 (previous step)

a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d ⊦ a0 = a, (a0 ∈ st → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ ¬ a ∈ st, a ∈ st ∧ st[a] = d ∧ (a1 ∈ st -- a → a1 ∈ st0 -- a ∧ st[a1] = st0[a1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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