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Proof step 6 from proof Diff-Diff-com (previous step)

⊦ (a ∈ st \ s ∧ ¬ a ∈ s0 ↔ a ∈ (st \ s0) \ s) ∧ (a ∈ st \ s ∧ ¬ a ∈ s0 → ((st \ s) \ s0)[a] = ((st \ s0) \ s)[a])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ (a ∈ st \ s ∧ ¬ a ∈ s0 ↔ a ∈ (st \ s0) \ s) ∧ (a ∈ st \ s ∧ ¬ a ∈ s0 → (st \ s)[a] = ((st \ s0) \ s)[a])

Rule argument

Additional information

Used simplifier rules: Diff-get: ⊦ ¬ a ∈ s → (st \ s)[a] = st[a];