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Proof step 5 from proof subset-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ st, a ∈ st0, a ∈ st[a, st0[a]], ∀ a0. a0 ∈ st[a, st0[a]] → a0 ∈ st0 ∧ st[a, st0[a]][a0] = st0[a0] ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

Rule argument

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;