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Proof step 30 from proof mapex-new (previous step)

P(H0), ∀ r0. (r0 ∈ H ++ r ↔ r0 ∈ (∅ ++ r)[r, o] ∪ H0) ∧ (r0 ∈ H ++ r → (H ++ r)[r0] = ((∅ ++ r)[r, o] ∪ H0)[r0]) ⊦ r = null, P(H), r ∈ H0, r ∈ H, ∀ r. (r ∈ H0 ↔ r ∈ H) ∧ (r ∈ H0 → H0[r] = H[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
P(H0), ∀ r1. (r1 = r ∨ r1 ∈ H ↔ r1 ∈ (∅ ++ r)[r, o] ∪ H0) ∧ (r1 = r ∨ r1 ∈ H → (H ++ r)[r1] = ((∅ ++ r)[r, o] ∪ H0)[r1]) ⊦ r = null, P(H), r ∈ H, r ∈ H0, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)