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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

Proof step 7:
i < 0 ⊦ i→n(i) = (* m) ↔ ~ i = n→i(m)
Proof step 8:
~(2 * i) = n→i(m0 * 2 + n0), i < 0, n0 < 2, 2 * i < 0 ⊦ m0 = i→n(i)

Rule argument

spec: <toplevel> 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: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as)%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