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Proof step 83 from proof embed-split (previous step)

mk(rs, f, g) = mk(rs0, f0, g0) ∪ mk(rs1, f1, g1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), r ∈ rs, r ∈ rs1 → g1(r) = a, r ∈ rs0 → g0(r) = a, ∀ r. r ∈ rs0 → g0(r) = a, ∀ r. r ∈ rs1 → g1(r) = a ⊦ # rs = 0, g(r) = a, rs = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 84:
⊦ Ha0 = Ha1 ↔ (∀ r. (r ∈ Ha0 ↔ r ∈ Ha1) ∧ (r ∈ Ha0 → Ha0[r]o = Ha1[r]o ∧ Ha0[r]a = Ha1[r]a))
Proof step 85:
rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), r ∈ rs, r ∈ rs0 → g0(r) = a, r ∈ rs1 → g1(r) = a, ∀ r. r ∈ rs1 → g1(r) = a, ∀ r. r ∈ rs0 → g0(r) = a, ∀ r. (r ∈ mk(rs, f, g) ↔ r ∈ mk(rs0, f0, g0) ∪ mk(rs1, f1, g1)) ∧ (r ∈ mk(rs, f, g) → mk(rs, f, g)[r]o = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]o ∧ mk(rs, f, g)[r]a = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]a) ⊦ # rs = 0, g(r) = a, rs = ∅

Rule argument

spec: heap, inst: heap-aux name: ext seq: ⊦ Ha0 = Ha1 ↔ (∀ r. (r ∈ Ha0 ↔ r ∈ Ha1) ∧ (r ∈ Ha0 → Ha0[r]o = Ha1[r]o ∧ Ha0[r]a = Ha1[r]a)) subst: [Ha1, Ha0] → [mk(rs0, f0, g0) ∪ mk(rs1, f1, g1), mk(rs, f, g)] flags: <all paths>

Additional information

Used terms: mk(rs0, f0, g0) ∪ mk(rs1, f1, g1), mk(rs, f, g)