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Proof step 3 from proof n2s-chars (previous step)

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'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 10 ≤ n → n→s(n) = n→s(n / 10) + n→s(n % 10)
Proof step 5:
¬ 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'
Proof step 51:
10 ≤ m, char ∈ s→l(n→s(m / 10) + 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'

Rule argument

spec: string2nat inst: name: n2s-rec seq: ⊦ 10 ≤ n → n→s(n) = n→s(n / 10) + n→s(n % 10) subst: [n] → [m] flags: rewrite, current

Additional information

Used terms: m