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Proof step 66 from proof embed-split (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), ∀ 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])

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ rs → mk(rs, f, g)[r]a = g(r); (s)% ⊦ r ∈ rs → mk(rs, f, g)[r]o = f(r); (s)% ⊦ r ∈ mk(rs, f, g) ↔ r ∈ rs; (s)