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Proof step 36 from proof size-exgr (
previous step
)
# s1 + 1 < # s0, # s1 < # s0, a0 ∈ s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s0, a ∈ s1, a0 ∈ s1, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1
Heuristic quantifier closing applied rule exists right
And closed the goal!
Rule argument
pos: right 5 terms: a0
Additional information
Used terms: a0