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Proof step 35 from proof lift-unknown (previous step)

rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1, ∀ r1. (r1 ∈ rs1 → r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ f(r);(∅ ++ r)[r1]o) ∧ g(r1) = (r1 = r ⊃ g(r);(∅ ++ r)[r1]a)) ⊦ # rs1 = 0, mk(rs1, f) = (∅ ++ r)[r, f(r)], r = null, rs1 = ∅, ∃ o. ((r => o))(mk(rs1, f)), ∃ o. (g(r)[r => o] * R)[rs1 ∪ rs0, f, g]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 37:
rs1 ∩ rs0 = ∅, R[rs0, f, g], r ∈ rs1, ∀ r1. (r1 ∈ rs1 → r1 = r) ∧ (r1 ∈ rs1 → f(r1) = (r1 = r ⊃ f(r);(∅ ++ r)[r1]o) ∧ g(r1) = (r1 = r ⊃ g(r);(∅ ++ r)[r1]a)) ⊦ # rs1 = 0, r = null, rs1 = ∅, ∀ r0. (r0 ∈ mk(rs1, f) ↔ r0 ∈ (∅ ++ r)[r, f(r)]) ∧ (r0 ∈ mk(rs1, f) → mk(rs1, f)[r0] = (∅ ++ r)[r, f(r)][r0]), ∃ o. (g(r)[r => o] * R)[rs1 ∪ rs0, f, g], ∃ o. ((r => o))(mk(rs1, f))

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)