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Proof step 16 from proof size-exgr (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
a ∈ s0, ∀ s. # s ≠ 0 → (∃ a0. a0 ∈ s) ⊦ # s0 = 0, # s0 = 1, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a

Rule argument

Additional information

Used simplifier rules: Size-empty: ⊦ # ∅ = 0;%In-empty: ⊦ ¬ a ∈ ∅;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2