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Proof step 58 from proof natstring-chars (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 59:
10 ≤ m, char ∈ n→s(m / 10) ∨ char ∈ n→s(m % 10), ∀ m1, char. m1 < m → char ∈ n→s(m1) → char = '0' ∨ char = '9' ∨ char = '8' ∨ char = '7' ∨ char = '6' ∨ char = '5' ∨ char = '4' ∨ char = '3' ∨ char = '2' ∨ char = '1' ⊦ char ∈ n→s(m % 10), char = '1', char = '2', char = '3', char = '4', char = '5', char = '6', char = '7', char = '8', char = '9', char = '0', m = 0

Rule argument

subst: [char, m1] → [char, m % 10]

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)%n ≠ 0 ⊦ ¬ n ≤ m % n; Used terms: char, m % 10