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Proof step 6 from proof Diff-del-rw (previous step)

⊦ (a0 ∈ st -- a ∧ ¬ a0 ∈ s ↔ a0 ∈ (st \ s) -- a) ∧ (a0 ∈ st -- a ∧ ¬ a0 ∈ s → ((st -- a) \ s)[a0] = (st \ s)[a0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ a0 ∈ st -- a ∧ ¬ a0 ∈ s ↔ a0 ∈ (st \ s) -- a

Rule argument

Additional information

Used simplifier rules: Diff-get: ⊦ ¬ a ∈ s → (st \ s)[a] = st[a];%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%Del-notin: ⊦ ¬ a ∈ st -- a; from specification store