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Proof step 6 from proof disj (previous step)

∀ r. ¬ r ∈ rs0 ∨ ¬ r ∈ rs1 ⊦ #(rs0 ∩ rs1) = 0, ∀ r. r ∈ rs0 ∩ rs1 ↔ r ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
r ∈ rs1, r ∈ rs0, ∀ r. ¬ r ∈ rs0 ∨ ¬ r ∈ rs1 ⊦ #(rs0 ∩ rs1) = 0, # rs1 = 0, # rs0 = 0, rs0 = ∅, rs1 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]%int-c: ⊦ rs ∩ rs0 = rs0 ∩ rs; from specification set-union[refset]%int-a: ⊦ (rs ∩ rs0) ∩ rs1 = rs ∩ rs0 ∩ rs1; from specification set-union[refset]% ⊦ r ∈ rs ∩ rs0 ↔ r ∈ rs ∧ r ∈ rs0; (ws)% ⊦ ¬ r ∈ ∅; (s)