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Proof step 3 from proof del (previous step)

⊦ ∀ r0. (r0 ∈ mk(rs -- r, f) ↔ r0 ∈ mk(rs, f) -- r) ∧ (r0 ∈ mk(rs -- r, f) → mk(rs -- r, f)[r0] = (mk(rs, f) -- r)[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r0 ∈ rs ⊦ # rs = 0, mk(rs -- r, f)[r0] = f(r0), r0 = r, rs = ∅

Rule argument

Additional information

Used simplifier rules: in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs;%at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r);%size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]%r ∈ H ⊦ r ≠ r0 → (H -- r0)[r] = H[r]; (s)% ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H; (s)% ⊦ r ∈ rs -- r0 ↔ r ≠ r0 ∧ r ∈ rs; (s)