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

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

User applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 3 terms: s0 abort: F discard: T computed: T

Additional information

Used terms: s0