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Proof step 11 from proof union-size-le (previous step)

a ∈ s ∪ s0, # s0 + # s + 1 < #(s ∪ s0), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ #(s ∪ s0) = 0, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ s0, # s0 + # s + 1 < #(s ∪ s0), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ s0 = ∅, s ∪ s0 = ∅, #(s ∪ s0) = 0, # s0 = 0, a ∈ s

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic