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Proof step 33 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 ∈ H, r ∈ H0, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 34:
r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])
Proof step 35:
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]), r0 ∈ (∅ ++ r)[r, o] ∪ H0 ∧ (∅ ++ r)[r, o] ⊥ H0
Proof step 38:
r0 = r ∨ r0 ∈ H ↔ r0 ∈ (∅ ++ r)[r, o] ∪ H0, P(H0), r0 ∈ (∅ ++ r)[r, o] ∪ H0, (∅ ++ r)[r, o] ⊥ H0, r0 = r ∨ r0 ∈ H → (H ++ r)[r0] = (r0 ∈ (∅ ++ r)[r, o] ⊃ (∅ ++ r)[r, o][r0];H0[r0]) ⊦ r = null, P(H), r ∈ H0, r ∈ H, (r0 ∈ H0 ↔ r0 ∈ H) ∧ (r0 ∈ H0 → H0[r0] = H[r0])

Rule argument

spec: heap name: at-uni seq: r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]) subst: [r, H1, H0] → [r0, H0, (∅ ++ r)[r, o]] flags: <all paths>

Additional information

Used terms: r0, H0, (∅ ++ r)[r, o]