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Proof step 6 from proof n2s-chars (previous step)

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'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n < 10 → n→s(n) = ('0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9'))[n] + ""
Proof step 8:
m < 10, char ∈ s→l(('0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9'))[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

spec: string2nat inst: name: n2s-char seq: ⊦ n < 10 → n→s(n) = ( '0' + '1' + '2' + '3' + '4' + '5' + '6' + '7' + ('8' + '9') )[n ] + "" subst: [n] → [m] flags: rewrite, current

Additional information

Used terms: m