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Proof step 1 from proof eq-02 (conclusion)

⊦ ¬ a ∈ s → (s ++ a = ∅ ++ b ↔ a = b ∧ s = ∅)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: one-size: ⊦ s = ∅ ++ a → # s = 1;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%in-05: ⊦ s ++ a = s0 ++ b ∧ a ≠ b → a ∈ s0;%size-zero-02: ⊦ # s = 0 → s = ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0;%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