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Proof step 7 from proof exin-uni (previous step)

∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0 ⊦ r = null, ∃ H0, o. H = H0 ∪ (∅ ++ r)[r, o] ∧ ¬ r ∈ H0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ r = null, r ∈ H0, ∃ H1, o0. (H0 ++ r)[r, o] = H1 ∪ (∅ ++ r)[r, o0] ∧ ¬ r ∈ H1

Rule argument

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