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Proof step 1 from proof Data-in-delput (conclusion)

⊦ ¬ d0 ∈ st ∧ d0 ≠ d1 → ¬ d0 ∈ st[b, d1] -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ st[b, d1] ⊦ a0 = a, st[b, d1][a0] = d1, ∃ a. a ∈ st ∧ st[a] = st[b, d1][a0]

Rule argument

Additional information

Used simplifier rules: Data-in: ⊦ d ∈ st ↔ (∃ a. a ∈ st ∧ st[a] = d);%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a]; from specification store%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; from specification store