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Proof step 9 from proof size-two-06 (previous step)

#(s0 ++ a0) < 2, a ∈ s0, ∀ a, b. # s0 < 2 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ b = a

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: b, a

Additional information

Used simplifier rules: size-less-01: ⊦ ¬ #(s ++ a) < # s;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%size-zero-02: ⊦ # s = 0 → s = ∅;%Size-empty: ⊦ # ∅ = 0;%In-empty: ⊦ ¬ a ∈ ∅;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s)% ⊦ m < 1 ↔ m = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic Used terms: b, a