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Proof step 10 from proof subseteq-inter-empty-empty (previous step)

a ∈ s2, a ∈ s0, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2, ∀ a. a ∈ s0 → a ∈ s1 ⊦ s2 = ∅, s0 = ∅, #(s0 ∩ s2) = 0, # s2 = 0, # s0 = 0

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 11:
a ∈ s2, a ∈ s0, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2, a ∈ s0 → a ∈ s1, ∀ a. a ∈ s0 → a ∈ s1 ⊦ s2 = ∅, s0 = ∅, #(s0 ∩ s2) = 0, # s2 = 0, # s0 = 0

Rule argument

pos: left 4 terms: a

Additional information

Used terms: a