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Proof step 5 from proof union-size-ls-in (previous step)

a1 ∈ s1, a0 ∈ s0, a ∈ s1, a ∈ s0, # s0 + # s1 ≤ # s0 + # s1 - #(s0 ∩ s1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
#(s0 ∩ s1) = 0, a1 ∈ s1, a0 ∈ s0, a ∈ s1, a ∈ s0, ∀ a2. ¬ a2 ∈ s0 ∨ ¬ a2 ∈ s1 ⊦

Rule argument

Additional information

Used simplifier rules: intersection-size: ⊦ ¬ # s0 + # s < #(s0 ∩ s);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic% ⊦ m + n ≤ m ↔ n = 0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)