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Proof step 6 from proof n2s-empty (previous step)

¬ n < 10, n→s(n) = "", ∀ n0. n0 < n → n→s(n0) = "" → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n→s(n / 10) = "", n→s(n % 10) = "", 10 ≤ n, ∀ n0. n0 < n → n→s(n0) = "" → false ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: n2s-rec: ⊦ 10 ≤ n → n→s(n) = n→s(n / 10) + n→s(n % 10);%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ str1 + str2 = "" ↔ str1 = "" ∧ str2 = "";% ⊦ ¬ m < n ↔ n ≤ m;