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Proof step 4 from proof union-left-singleton (previous step)

r ∈ H ⊦ r = null, ∀ r0. (r0 ∈ (∅ ++ r)[r, H[r]] ∪ (H -- r) ↔ r0 ∈ H) ∧ (r0 ∈ (∅ ++ r)[r, H[r]] ∪ (H -- r) → ((∅ ++ r)[r, H[r]] ∪ (H -- r))[r0] = H[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
r ∈ H ⊦ (r0 ∈ (∅ ++ r)[r, H[r]] ∪ (H -- r) ↔ r0 ∈ H) ∧ (r0 ∈ (∅ ++ r)[r, H[r]] ∪ (H -- r) → ((∅ ++ r)[r, H[r]] ∪ (H -- r))[r0] = H[r0])

Rule argument

Additional information

Used simplifier rules: nonnull: ⊦ r ∈ H → r ≠ null;