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Proof step 17 from proof abs-any-def (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 18:
i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ i→n(i / 10) < i→n(i), j = 0, abs(mkDecimal(i / 10, abs(j) - 1)) = mkDecimal(abs(i), j)
Proof step 23:
abs(mkDecimal(i / 10, abs(j) - 1)) = mkDecimal(abs(i / 10), abs(j) - 1), i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ j = 0, abs(mkDecimal(i / 10, abs(j) - 1)) = mkDecimal(abs(i), j)

Rule argument

subst: [i0, j] → [i / 10, abs(j) - 1]

Additional information

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