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Proof step 14 from proof diff-sub-size (
previous step
)
a0 ∈ s0, a ∈ s0, a1 ∈ s, ∀ a2. a2 ∈ s → a2 ∈ s0 ⊦ s = ∅, #(s0 \ s) - 1 = # s0 - # s + 1, # s = 0, a ∈ s, a1 ∈ s0
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 4 terms: a1
Additional information
Used terms: a1