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Proof step 2 from proof exin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
r ∈ H ⊦ ∃ H0, o. H = (H0 ++ r)[r, o] ∧ ¬ r ∈ H0

Rule argument

Additional information

Used simplifier rules: in-write: ⊦ r ∈ H[r0, o] ↔ r ∈ H;%in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;%nonnull: ⊦ r ∈ H → r ≠ null;