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Proof step 1 from proof Data-notin-other-3 (conclusion)

⊦ d ≠ d0 ∧ d0 ≠ d1 ∧ d0 ≠ d2 ∧ ¬ d0 ∈ st → ¬ d0 ∈ st[a, d1][b, d2][c, d]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ st[a, d1][b, d2][c, d] ⊦ d = st[a, d1][b, d2][c, d][a0], st[a, d1][b, d2][c, d][a0] = d1, st[a, d1][b, d2][c, d][a0] = d2, ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][b, d2][c, d][a0]

Rule argument

Additional information

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