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Proof step 15 from proof mapex-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
P(H), r0 ∈ (∅ ++ r)[r, (H ++ r)[r]] ∪ H ⊦ r = null, r0 = r, r ∈ H, r0 ∈ H

Rule argument

Additional information

Used simplifier rules: r ≠ null ⊦ (∅ ++ r)[r, o] ⊥ H ↔ ¬ r ∈ H; (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)% ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ H[r0, o] ↔ r ∈ H; (s)% ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1); (s)