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Proof step 20 from proof neg-any-def (previous step)

i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → mkDecimal(~ i0, j) = ~ mkDecimal(i0, j) ⊦ j = 0, mkDecimal(~(i / 10), abs(j) - 1) = ~ mkDecimal(i / 10, abs(j) - 1)

User applied rule apply induction

And obtained the following new goals

Proof step 21:
i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → mkDecimal(~ i0, j) = ~ mkDecimal(i0, j) ⊦ i→n(i / 10) < i→n(i), j = 0, mkDecimal(~(i / 10), abs(j) - 1) = ~ mkDecimal(i / 10, abs(j) - 1)

Rule argument

subst: [i0, j] → [i / 10, abs(j) - 1]

Additional information

Used terms: i / 10, abs(j) - 1