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Proof step 10 from proof size-one-02 (previous step)

#(s0 ++ a0) = 1, a ∈ s0, ∀ a, b. # s0 = 1 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ a = b, a0 ∈ s0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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