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Proof step 7 from proof union-left-singleton (previous step)

r ∈ H, (∅ ++ r)[r, H[r]] ⊥ H -- r ⊦ (r0 ∈ (∅ ++ r)[r, H[r]] ∨ r0 ∈ H -- r ↔ r0 ∈ H) ∧ (r0 ∈ (∅ ++ r)[r, H[r]] ∨ r0 ∈ H -- r → ((∅ ++ r)[r, H[r]] ∪ (H -- r))[r0] = H[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
r ∈ H, r0 ∈ H ⊦ ((∅ ++ r)[r, H[r]] ∪ (H -- r))[r0] = H[r0]

Rule argument

Additional information

Used simplifier rules: nonnull: ⊦ r ∈ H → r ≠ null;%disj-singleton-write: r ≠ null ⊦ (∅ ++ r)[r, o] ⊥ H ↔ ¬ r ∈ H;%in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;%in-write: ⊦ r ∈ H[r0, o] ↔ r ∈ H;%in-emp: ⊦ ¬ r ∈ ∅;%in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;