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

⊦ ¬ a ∈ st → (d0 ∈ st[a, d1] ↔ d0 ∈ st ∨ d0 = d1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ st ∧ st[a0] = d0 ∨ d0 = d1 ⊦ a ∈ st, ∃ a0. a0 ∈ st[a, d1] ∧ st[a, d1][a0] = d0

Rule argument

Additional information

Used simplifier rules: Data-notin-other: ⊦ d0 ≠ d1 ∧ ¬ d0 ∈ st → ¬ d0 ∈ st[a, d1];%Data-in: ⊦ d ∈ st ↔ (∃ a. a ∈ st ∧ st[a] = d);