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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, ∀ a0. ¬ a0 ∈ s1 ++ a ∨ ¬ a0 ∈ s0 ⊦ #((s0 ∪ s1) ++ a) = #(s1 ++ a) + # s0

Rule argument

Additional information

Used simplifier rules: unadd-01: ⊦ s ∪ (s0 ++ a) = (s ∪ s0) ++ a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;