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Proof step 3 from proof intersection-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# s0 + #(s1 ++ a) < #(s0 ∩ (s1 ++ a)), ∀ s0. # s0 + # s1 < #(s0 ∩ s1) → #(s0 ∩ s1) = 0 ∨ s0 ∩ s1 = ∅ ⊦ #(s0 ∩ (s1 ++ a)) = 0, s0 ∩ (s1 ++ a) = ∅

Rule argument

Additional information

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