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Proof step 54 from proof n2s-chars (previous step)

10 ≤ m, char ∈ s→l(n→s(m % 10)), ∀ m1, char. m1 < m → true → ¬ char ∈ '0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9') → ¬ char ∈ s→l(n→s(m1)) ⊦ char = '0', char = '9', char = '8', char = '7', char = '6', char = '5', char = '4', char = '3', char = '2', char = '1', m = 0, char ∈ s→l(n→s(m / 10))

User applied rule apply induction

And closed the goal!

Rule argument

subst: [char, m1] → [char, m % 10]

Additional information

Used simplifier rules: n ≠ 0 ⊦ ¬ n ≤ m % n;% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ char + char0 = char ' + char0 ';% ⊦ char + chars = char ' + chars;% ⊦ char ≠ char0 → (char ∈ char0 ' + chars ↔ char ∈ chars);% ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1; Used terms: char, m % 10