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Proof step 34 from proof size-exgr (previous step)

# s1 + 1 < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, s0 = ∅, a ∈ s1, a ∈ s0, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 35:
# s1 + 1 < # s0, # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1), ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, s0 = ∅, a ∈ s1, a ∈ s0, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1

Rule argument

pos: left 2 terms: s0

Additional information

Used terms: s0