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Proof step 8 from proof size-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ s1, ∀ a0. ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0, ∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0 ⊦ s1 = ∅, #(s0 ∪ s1) = # s1 + # s0, # s1 = 0

Rule argument

Additional information

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