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Proof step 116 from proof write-known (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a, ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → (r = r0 ⊃ o1;f(r0)) = (r0 = r ⊃ o1;(∅ ++ r)[r0])) ⊦ mk(rs1, f) = (∅ ++ r)[r, f(r)], r = null, (a[r => f(r)] * R)[rs1 ∪ rs0, f, g], r ∈ rs0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 117:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 118:
rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r0. (r0 ∈ rs1 ↔ r0 = r) ∧ (r0 ∈ rs1 → (r = r0 ⊃ o1;f(r0)) = (r0 = r ⊃ o1;(∅ ++ r)[r0])), ∀ r. r ∈ rs1 → g(r) = a ⊦ r = null, (a[r => f(r)] * R)[rs1 ∪ rs0, f, g], r ∈ rs0, ∀ r0. (r0 ∈ mk(rs1, f) ↔ r0 ∈ (∅ ++ r)[r, f(r)]) ∧ (r0 ∈ mk(rs1, f) → mk(rs1, f)[r0] = (∅ ++ r)[r, f(r)][r0])

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, f(r)], mk(rs1, f)] flags: <all paths>

Additional information

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