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Proof step 2 from proof Data-in-delput-2 (previous step)

d0 ∈ st[b, d1][c, d2] -- a ⊦ d0 = d2, d0 = d1, d0 ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 ∈ st[b, d1][c, d2] ⊦ st[b, d1][c, d2][a0] = d2, st[b, d1][c, d2][a0] = d1, a0 = a, ∃ a. a ∈ st ∧ st[a] = st[b, d1][c, d2][a0]

Rule argument

Additional information

Used simplifier rules: Data-in: ⊦ d ∈ st ↔ (∃ a. a ∈ st ∧ st[a] = d);% ⊦ a ≠ b → (st -- b)[a] = st[a]; (s)% ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (ws)