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Proof step 3 from proof Data-in-delput-2 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a0 ∈ st[b, d1] ⊦ a0 = c, a0 = a, st[b, d1][a0] = d1, st[b, d1][a0] = d2, ∃ a1. a1 ∈ st ∧ st[a1] = st[b, d1][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