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Proof step 22 from proof aux-known (previous step)

mk(rs1, f) = (∅ ++ r)[r, o], rs1 ∩ rs0 = ∅, R[rs0, f, g], r0 ∈ rs1 → g(r0) = a0, ∀ r. r ∈ rs1 → g(r) = a0 ⊦ r = null, (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)], (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 24:
rs1 ∩ rs0 = ∅, R[rs0, f, g], r0 ∈ rs1 → g(r0) = a0, ∀ r. r ∈ rs1 → g(r) = a0, ∀ r0. (r0 ∈ mk(rs1, f) ↔ r0 ∈ (∅ ++ r)[r, o]) ∧ (r0 ∈ mk(rs1, f) → mk(rs1, f)[r0] = (∅ ++ r)[r, o][r0]) ⊦ r = null, (a1[r => o] * R)[rs1 ∪ rs0, f, g(r; a1)] ∧ g(r) = a0, (r0 ∈ rs1 → (r = r0 ⊃ a1;g(r0)) = a1) ∧ R[rs0, f, g(r; a1)]

Rule argument

spec: heap name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [(∅ ++ r)[r, o], mk(rs1, f)] flags: <all paths>

Additional information

Used terms: (∅ ++ r)[r, o], mk(rs1, f)