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Proof step 4 from proof dec-int-01 (previous step)

0 ≤ i ⊦ encode(2 * i→n(i)) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n % 2 = 0 → encode(n) = n→i(n / 2)

Rule argument

spec: <toplevel> name: enc-nat-pos seq: ⊦ n % 2 = 0 → encode(n) = n→i(n / 2) subst: [n] → [2 * i→n(i)] flags: <all paths>

Additional information

Used simplifier rules: enc-nat-neg: ⊦ n % 2 ≠ 0 → encode(n) = ~ n→i((n + 1) / 2);%n ≠ 0 ⊦ (m * n) % n = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ (2 * m) / 2 = m; (s)% ⊦ 0 ≤ i → n→i(i→n(i)) = i; (s)% ⊦ (2 * m + 1) / 2 = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ i + i = 2 * i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 2 * i = 0 ↔ i = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 * i→n(i)