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Proof step 7 from proof diff-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
r0 ∈ H1, ∀ H, r, H0. # H < # H1 → true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1

Rule argument

Additional information

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