Proof step 1 from proof log2ub-log2 (conclusion) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 2:

  ⊦ log2↑(n) = (∃ m. n = 2^ m ⊃ log2 n;log2(2 * n + 1))

• Proof step 3:

  ⊦ (∃ m. n = 2^ m ⊃ log2 n;log2(2 * n + 1)) = (n = 0 ⊃ 0;(∃ m. n = 2^ m ⊃ log2 n;log2 n + 1))

Rule argument

Additional information

Used terms: n

Proof step 2 from proof log2ub-log2 (previous step) (back to tree)

Lemma log2up-def

Proof step 3 from proof log2ub-log2 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

  ∀ m. n ≠ 2^ m ⊦ log2(2 * n + 1) = (n = 0 ⊃ 0;(∃ m. n = 2^ m ⊃ log2 n;log2 n + 1))

Rule argument

Additional information

Used simplifier rules: ⊦ 2^ n ≠ 0; (s) ⊦ 2^ n = 2^ m ↔ n = m; (s) ⊦ log2 2^ n = n; (s)

Proof step 4 from proof log2ub-log2 (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 5:

  n ≠ 0, ∀ m. n ≠ 2^ m ⊦ log2(2 * n + 1) = (∃ m. n = 2^ m ⊃ log2 n;log2 n + 1)

Rule argument

Additional information

Used simplifier rules: mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ log2 1 = 0; (s)

Proof step 5 from proof log2ub-log2 (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  n ≠ 0, ∀ m. n ≠ 2^ m ⊦ log2 n + 1 = (∃ m. n = 2^ m ⊃ log2 n;log2 n + 1)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ n ≠ 0 → log2(2 * n + 1) = log2 n + 1; (s)

Proof step 6 from proof log2ub-log2 (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ log2 2^ n = n; (s) ⊦ m = m + n ↔ n = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ 2^ n ≠ 0; (s) ⊦ 2^ n = 2^ m ↔ n = m; (s)