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Proof step 16 from proof intersection-size (previous step)

a ∈ s1, # s0 + # s1 < #(s0 ∩ s1), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #(s0 ∩ s1) = 0, s0 ∩ s1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# s0 + # s1 < #(s0 ∩ s1), a ∈ s1, ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ s1 = ∅, s0 ∩ s1 = ∅, #(s0 ∩ s1) = 0, # s1 = 0

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1