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

char ∈ n→s(m) ⊦ char = '0', char = '9', char = '8', char = '7', char = '6', char = '5', char = '4', char = '3', char = '2', char = '1'

User applied rule induction

And obtained the following new goals

Proof step 6:
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 = '0' ∨ char = '9' ∨ char = '8' ∨ char = '7' ∨ char = '6' ∨ char = '5' ∨ char = '4' ∨ char = '3' ∨ char = '2' ∨ char = '1'

Rule argument

indvar: m wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: m