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Proof step 3 from proof disj (previous step)

⊦ (∀ r. ¬ r ∈ mk(rs0, f0) ∨ ¬ r ∈ mk(rs1, f1)) ↔ rs0 ∩ rs1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ r. ¬ r ∈ rs0 ∨ ¬ r ∈ rs1 ⊦ #(rs0 ∩ rs1) = 0, rs0 ∩ rs1 = ∅

Rule argument

Additional information

Used simplifier rules: in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs;%size-zero: ⊦ # rs ≠ 0 → rs ≠ ∅; (forward) from specification set-basic[refset]%size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]% ⊦ rs ∩ rs0 = ∅ ∧ r ∈ rs → (r ∈ rs0 ↔ false); (as)%int-c: ⊦ rs ∩ rs0 = rs0 ∩ rs; from specification set-union[refset]%int-a: ⊦ (rs ∩ rs0) ∩ rs1 = rs ∩ rs0 ∩ rs1; from specification set-union[refset]