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Proof step 14 from proof size-exgr (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
a ∈ s0, # s1 + 1 < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s1, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;