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Proof step 6 from proof sw-del-add-new (previous step)

⊦ r = null, r ∈ H, r ∈ rs, (r0 = r ∨ r0 ∈ H ∧ ¬ r0 ∈ rs → (r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs) ∧ ((r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs → ((H ++ r)[r, o] \ rs)[r0] = (r0 = r ⊃ o;((H \ rs) ++ r)[r0]))

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
r0 = r ⊦ r = null, r ∈ rs, r ∈ H, (r0 = r ∨ r0 ∈ H ∧ ¬ r0 ∈ rs → (r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs) ∧ ((r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs → ((H ++ r)[r, o] \ rs)[r0] = (r0 = r ⊃ o;((H \ rs) ++ r)[r0]))
Proof step 10:
⊦ r = null, r0 = r, r ∈ rs, r ∈ H, (r0 = r ∨ r0 ∈ H ∧ ¬ r0 ∈ rs → (r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs) ∧ ((r0 = r ∨ r0 ∈ H) ∧ ¬ r0 ∈ rs → ((H ++ r)[r, o] \ rs)[r0] = (r0 = r ⊃ o;((H \ rs) ++ r)[r0]))

Rule argument

fma: r0 = r

Additional information

Used formulas: r0 = r