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Proof step 13 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])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r])
Proof step 15:
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
Proof step 18:
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]))

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, H, (∅ ++ r)[r, (H ++ r)[r]]] flags: <all paths>

Additional information

Used terms: r0, H, (∅ ++ r)[r, (H ++ r)[r]]