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Proof step 8 from proof precise-mapex (previous step)

r ∈ (∅ ++ r)[r, o] ∪ H0 ↔ r ∈ (∅ ++ r)[r, o0] ∪ H, r ∈ (∅ ++ r)[r, o] ∪ H0 → ((∅ ++ r)[r, o] ∪ H0)[r] = ((∅ ++ r)[r, o0] ∪ H)[r] ⊦ (∅ ++ r)[r, o] = (∅ ++ r)[r, o0], r = null, r ∈ H, r ∈ H0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ (∅ ++ r)[r, o] = (∅ ++ r)[r, o0], r = null, r ∈ H0, r ∈ H, r ∈ (∅ ++ r)[r, o] ∪ H0, r ∈ (∅ ++ r)[r, o0] ∪ H

Rule argument

Additional information

Used simplifier rules: r ≠ null, ¬ r ∈ H ⊦ ((∅ ++ r)[r, o] ∪ H)[r] = o; (s)