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Proof step 3 from proof Diff-del-rw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (a0 ∈ (st -- a) \ s ↔ a0 ∈ (st \ s) -- a) ∧ (a0 ∈ (st -- a) \ s → ((st -- a) \ s)[a0] = (st \ s)[a0])

Rule argument

Additional information

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