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Proof step 8 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]), ∀ r. (r ∈ H1 ↔ r ∈ ∅) ∧ (r ∈ H1 → H1[r] = (∅)[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
r0 ∈ H1, 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])

Rule argument

Additional information

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