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Proof step 9 from proof getint-pos (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0, j] → [i / 10, abs(j) - 1]

Additional information

Used simplifier rules: lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%div-zero-ngr: ⊦ i < 0 ∧ 0 < j → ¬ 0 < i / j; from specification int-div%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%div-lezero: ⊦ 0 < j → (i / j ≤ 0 ↔ i < j); from specification int-div%mod-zero-08: 0 < j ⊦ 0 < i ∧ i < j → i % (* j) = i; from specification int-div%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%abs-div10-le: ⊦ abs(i) ≤ abs(i) / 10 ↔ i = 0; from specification int-div Used terms: i / 10, abs(j) - 1