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Proof step 4 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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
∀ H, r, H0. # H < # H1 → true → (r ∈ H0 \ H ↔ r ∈ H0 ∧ ¬ r ∈ H) ⊦ r ∈ H0 \ H1 ↔ r ∈ H0 ∧ ¬ r ∈ H1, H1 = ∅

Rule argument

fma: H1 = ∅

Additional information

Used simplifier rules: diff-emp: ⊦ H0 \ ∅ = H0;%in-emp: ⊦ ¬ r ∈ ∅; Used formulas: H1 = ∅