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Proof step 34 from proof diff-sub-size (previous step)

a0 ∈ s, ∀ a1. a1 = a ∨ a1 ∈ s → a1 ∈ s0 ⊦ s = ∅, #(s0 \ s) = # s0 - #(s ++ a), # s = 0, a ∈ s0 \ s, a0 ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0

Additional information

Used terms: a0