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Proof step 9 from proof size-one (previous step)

# s = 1, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%size-zero-02: ⊦ # s = 0 → s = ∅;%Size-empty: ⊦ # ∅ = 0;%In-empty: ⊦ ¬ a ∈ ∅;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2