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Proof step 9 from proof natstring-chars (previous step)

¬ 10 ≤ m, char ∈ n→s(m), ∀ 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 10:
char ∈ n→s(m), 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'

Rule argument

Additional information

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