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Proof step 16 from proof modify-sym (previous step)

⊦ r = null, r ∈ H0, r ∈ rs, ∃ r0. ¬ r0 ∈ rs ∧ r0 ≠ null ∧ r0 ≠ r ∧ ¬ r0 ∈ H0 ∧ H0 = H0 ++ r ++ r0, ∃ r0, o. ¬ r0 ∈ rs ∧ (r0 = r ∨ r0 ∈ H0) ∧ H0 = (H0 ++ r)[r0, o], ∃ r0. ¬ r0 ∈ rs ∧ (r0 = r ∨ r0 ∈ H0) ∧ H0 = (H0 ++ r) -- r0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 6 terms: r

Additional information

Used simplifier rules: del-new: ¬ r ∈ H, r ≠ null ⊦ (H ++ r) -- r = H; from specification heap Used terms: r