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Proof step 31 from proof equality-any (previous step)

∀ i1, i2, j1, j0. i→n(i1) + i→n(i2) < i→n(i0 * 10 ^ abs(j)) + i→n(i0) → true → (mkDecimal(i1, j1) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j1)) ⊦ i0 % 10 = 0, (i0 * 10 ^ abs(j)) % 10 = 0, j = 0, i0 = 0, 10 ^ abs(j) % 10 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ 0 < i → j ^ i = j ^ i -1 * j

Rule argument

spec: int-pot inst: name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [abs(j), 10] flags: rewrite

Additional information

Used simplifier rules: sub-pred: ⊦ i -1 = i - 1; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%mod-zero-num: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0; from specification int-div%commutativity: ⊦ i * j = j * i; from specification int-mult%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs Used terms: abs(j), 10