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Proof step 20 from proof sep (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦ r ∈ rs0 ∪ rs1 ↔ r ∈ rs

Proof step 20 from proof sep (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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