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Proof step 30 from proof size-exgr (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
# s1 < # s0, a ∈ s1, a0 ∈ s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s0, a0 ∈ s1, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1

Rule argument

Additional information

Used simplifier rules: ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);