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Proof step 2 from proof Data-notin-other-3 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 ∈ st[a, d1][b, d2][c, d] ⊦ st[a, d1][b, d2][c, d][a0] = d2, st[a, d1][b, d2][c, d][a0] = d1, d = st[a, d1][b, d2][c, d][a0], ∃ 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);