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Proof step 10 from proof size-exgr (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# s1 < # s0, a ∈ s0, a ∈ s1, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1

Rule argument

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);