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Proof step 5 from proof Data-notin-other-2 (previous step)

a0 ∈ st[a, d1] ⊦ a0 = a, a0 = b, st[a, d1][a0] = d1, st[a, d1][a0] = d2, ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ st ⊦ a0 = a, a0 = b, st[a0] = d1, st[a0] = d2, ∃ a1. a1 ∈ st ∧ st[a1] = st[a0]

Rule argument

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); from specification store%At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; from specification store