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Proof step 5 from proof n2s-chars (previous step)

¬ 10 ≤ m, char ∈ s→l(n→s(m)), ∀ 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'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
char ∈ s→l(n→s(m)), 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'

Rule argument

Additional information

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