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Proof step 69 from proof embed-split (previous step)

r ∈ rs ↔ r ∈ mk(rs0, f0, g0) ∪ mk(rs1, f1, g1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), r ∈ rs → f(r) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]o ∧ g(r) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]a, ∀ r. r ∈ rs1 → g1(r) = a, ∀ r. r ∈ rs0 → g0(r) = a, ∀ r0. (r0 ∈ rs ↔ r0 ∈ mk(rs0, f0, g0) ∪ mk(rs1, f1, g1)) ∧ (r0 ∈ rs → f(r0) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r0]o ∧ g(r0) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r0]a) ⊦ (r ∈ rs ↔ r ∈ mk(rs0, f0) ∪ mk(rs1, f1)) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 70:
⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1)
Proof step 71:
r ∈ rs ↔ r ∈ mk(rs0, f0, g0) ∪ mk(rs1, f1, g1), rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), mk(rs0, f0) ⊥ mk(rs1, f1), r ∈ rs → f(r) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]o ∧ g(r) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]a, ∀ r0. (r0 ∈ rs ↔ r0 ∈ mk(rs0, f0, g0) ∪ mk(rs1, f1, g1)) ∧ (r0 ∈ rs → f(r0) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r0]o ∧ g(r0) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r0]a), ∀ r. r ∈ rs0 → g0(r) = a, ∀ r. r ∈ rs1 → g1(r) = a ⊦ (r ∈ rs ↔ r ∈ mk(rs0, f0) ∨ r ∈ mk(rs1, f1)) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r])

Rule argument

spec: heap name: in-uni seq: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1) subst: [H1, H0, r] → [mk(rs1, f1), mk(rs0, f0), r] flags: <all paths>

Additional information

Used simplifier rules: ⊦ mk(rs0, f0) ⊥ mk(rs1, f1) ↔ rs0 ∩ rs1 = ∅; (s) Used terms: mk(rs1, f1), mk(rs0, f0), r