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Proof step 3 from proof Data-in-putdel (previous step)

a0 ∈ (st -- a)[b, d1] ⊦ a0 = b, (st -- a)[b, d1][a0] = d1, ∃ a1. a1 ∈ st ∧ st[a1] = (st -- a)[b, d1][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a]; from specification store%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a]; from specification store%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; from specification store%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