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Proof step 16 from proof size-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
∀ s0. (∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s0) → #(s1 ∪ s0) = # s1 + # s0, ∀ a0. ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0 ⊦ #(s0 ∪ s1) = # s1 + # s0, a ∈ s0, a ∈ s1

Proof step 16 from proof size-union (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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