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Proof step 32 from proof mapex-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
r0 = r ∨ r0 ∈ H ↔ r0 ∈ (∅ ++ r)[r, o] ∪ H0, P(H0), r0 = r ∨ r0 ∈ H → (H ++ r)[r0] = ((∅ ++ r)[r, o] ∪ H0)[r0] ⊦ r = null, P(H), r ∈ H, r ∈ H0, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

Proof step 32 from proof mapex-new (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information