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Proof step 8 from proof diff-at (previous step)

r ∈ H0, ∀ H, r, H0. # H < # H1 → r ∈ H0 → (H0 \ H)[r] = H0[r] ∨ r ∈ H ⊦ (H0 \ H1)[r] = H0[r], r ∈ H1, ∀ r. (r ∈ H1 ↔ r ∈ ∅) ∧ (r ∈ H1 → H1[r] = (∅)[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
r ∈ H0, r0 ∈ H1, ∀ H, r, H0. # H < # H1 → r ∈ H0 → (H0 \ H)[r] = H0[r] ∨ r ∈ H ⊦ (H0 \ H1)[r] = H0[r], r ∈ H1

Rule argument

Additional information

Used simplifier rules: in-emp: ⊦ ¬ r ∈ ∅;