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Proof step 5 from proof at-uni (previous step)

H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1, ∀ H, H0, r. # H < # H1 → H0 ⊥ H ∧ (r ∈ H0 ∨ r ∈ H) → (H0 ∪ H)[r] = (r ∈ H0 ⊃ H0[r];H[r]) ⊦ (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])

User applied rule cut formula

And obtained the following new goals

Proof step 6:
H0 ⊥ H1, r ∈ H0 ∨ r ∈ H1, ∀ H, H0, r. # H < # H1 → H0 ⊥ H ∧ (r ∈ H0 ∨ r ∈ H) → (H0 ∪ H)[r] = (r ∈ H0 ⊃ H0[r];H[r]) ⊦ H1 = ∅, (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])

Rule argument

fma: H1 = ∅

Additional information

Used simplifier rules: uni-emp-r: ⊦ H ∪ ∅ = H;%disj-emp-r: ⊦ H ⊥ ∅;%in-emp: ⊦ ¬ r ∈ ∅; Used formulas: H1 = ∅