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Proof step 88 from proof embed-split (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 89:
⊦ Ha0 ⊥ Ha1 → (r ∈ Ha0 ∪ Ha1 ↔ r ∈ Ha0 ∨ r ∈ Ha1)
Proof step 90:
f(r) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]o, g(r) = (mk(rs0, f0, g0) ∪ mk(rs1, f1, g1))[r]a, rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), r ∈ rs, mk(rs0, f0, g0) ⊥ mk(rs1, f1, g1), r ∈ rs0 → g0(r) = a, r ∈ rs1 → 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 ∈ rs1 → g1(r) = a, ∀ r. r ∈ rs0 → g0(r) = a, r ∈ mk(rs0, f0, g0) ∨ r ∈ mk(rs1, f1, g1) ⊦ # rs = 0, g(r) = a, rs = ∅

Rule argument

spec: heap, inst: heap-aux name: in-uni seq: ⊦ Ha0 ⊥ Ha1 → (r ∈ Ha0 ∪ Ha1 ↔ r ∈ Ha0 ∨ r ∈ Ha1) subst: [Ha1, Ha0, r] → [mk(rs1, f1, g1), mk(rs0, f0, g0), r] flags: <all paths>

Additional information

Used simplifier rules: ⊦ mk(rs0, f, g) ⊥ mk(rs1, f0, g0) ↔ rs0 ∩ rs1 = ∅; (s) Used terms: mk(rs1, f1, g1), mk(rs0, f0, g0), r