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Proof step 55 from proof natstring-chars (previous step)

10 ≤ m, char ∈ n→s(m / 10) + 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'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
10 ≤ m, char ∈ n→s(m / 10) + 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

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)