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Proof step 6 from proof i2n-divtwo (previous step)

i→n(2 * i) = m0 * 2 + n0, n0 < 2, i < 0 ⊦ m0 = i→n(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: num-i2n-eq-01 seq: i < 0 ⊦ i→n(i) = (* m) ↔ ~ i = n→i(m) subst: [m, i] → [m0 * 2 + n0, 2 * i] flags: rewrite, current

Additional information

Used simplifier rules: zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: m0 * 2 + n0, 2 * i