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Proof step 10 from proof size-union (previous step)

a ∈ s1, ∀ a0. ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0, (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0 ⊦ s1 = ∅, #(s0 ∪ s1) = # s1 + # s0, # s1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a0 ∈ s0, a0 ∈ s1, a ∈ s1, ∀ a0. ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0 ⊦ s0 = ∅, # s1 = 0, #(s0 ∪ s1) = # s1 + # s0, s1 = ∅, # s0 = 0

Rule argument

Additional information

Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic