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Proof step 67 from proof embed-split (previous step)

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])

User applied rule all left

And obtained the following new goals

Proof step 68:
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, ∀ 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) ∪ mk(rs1, f1)) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r])

Rule argument

pos: left 6 terms: r abort: F discard: F computed: T

Additional information

Used terms: r