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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ (st -- a)[b, d1] ⊦ (st -- a)[b, d1][a0] = d1, ∃ a1. a1 ∈ st ∧ st[a1] = (st -- a)[b, d1][a0]

Rule argument

Additional information

Used simplifier rules: Data-in: ⊦ d ∈ st ↔ (∃ a. a ∈ st ∧ st[a] = d);