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Proof step 18 from proof mapex-new (previous step)

P(H), r0 ∈ (∅ ++ r)[r, (H ++ r)[r]] ∪ H, (∅ ++ r)[r, (H ++ r)[r]] ⊥ H ⊦ r = null, r ∈ H, (r0 = r ∨ r0 ∈ H ↔ r0 ∈ (∅ ++ r)[r, (H ++ r)[r]] ∪ H) ∧ (r0 = r ∨ r0 ∈ H → (H ++ r)[r0] = (r0 ∈ (∅ ++ r)[r, (H ++ r)[r]] ⊃ (∅ ++ r)[r, (H ++ r)[r]][r0];H[r0]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
P(H), r0 ∈ (∅ ++ r)[r, (H ++ r)[r]] ∪ H ⊦ r = null, r ∈ H, (r0 = r ∨ r0 ∈ H) ∧ (H ++ r)[r0] = (r0 = r ⊃ (H ++ r)[r];H[r0])

Rule argument

Additional information

Used simplifier rules: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]); (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)% ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ H[r0, o] ↔ r ∈ H; (s)%r ≠ null ⊦ (∅ ++ r)[r, o] ⊥ H ↔ ¬ r ∈ H; (s)