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Proof step 15 from proof intersection-size (previous step)

# s0 + # s1 + 1 < #(s0 ∩ s1) + 1, a0 ∈ s0, a ∈ s0, ∀ s0. ¬ # s0 + # s1 < #(s0 ∩ s1) ⊦ a ∈ s1, a ∈ s0 ∩ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# s0 + # s1 < #(s0 ∩ s1), a0 ∈ s0, a ∈ s0, ∀ s0. ¬ # s0 + # s1 < #(s0 ∩ s1) ⊦ a ∈ s1

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)