Proof step 1 from proof lzer (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(log2(m + 1) = 0 ↔ m = 0) ⊦

Rule argument

Additional information

Proof step 2 from proof lzer (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  log2(m + 1) = 0, m ≠ 0 ⊦

Rule argument

Additional information

Used simplifier rules: lone: ⊦ log2 1 = 0;

Proof step 3 from proof lzer (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 4:

  n ≠ 0 ⊦ ∃ m. 2^ m ≤ n ∧ n < 2^(m + 1)

• Proof step 5:

  log2(m + 1) = 0, m ≠ 0, ∃ m0. 2^ m0 ≤ m + 1 ∧ m + 1 < 2^(m0 + 1) ⊦

Rule argument

Additional information

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

Proof step 4 from proof lzer (previous step) (back to tree)

Lemma potex

Proof step 5 from proof lzer (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  log2(m + 1) = 0, m ≠ 0, 2^ m0 < 2^(m0 + 1), m + 1 < 2^(m0 + 1), 2^ m0 ≤ m + 1 ⊦

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 2^ n ≠ 0; fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 6 from proof lzer (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 7:

  ⊦ 2^ m ≤ n ∧ n < 2^ m +1 → log2 n = m

• Proof step 8:

  log2(m + 1) = 0, m ≠ 0, m + 1 < 2^(m0 + 1), 2^ m0 < 2^(m0 + 1), 2^ m0 ≤ m + 1, 2^ m0 ≤ m + 1 ∧ m + 1 < 2^ m0 +1 → log2(m + 1) = m0 ⊦

Rule argument

Additional information

Used terms: m0, m + 1

Proof step 7 from proof lzer (previous step) (back to tree)

Lemma Log2def

Proof step 8 from proof lzer (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  log2(m + 1) = 0, m ≠ 0, m + 1 < 2^ 1, 2^ 0 < 2^ 1, 2^ 0 ≤ m + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)

Proof step 9 from proof lzer (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ m ≤ n ↔ ¬ n < m

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ 2^ 1 = 2; n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s) ⊦ m < 1 ↔ m = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: m + 1, 2^ 0

Proof step 10 from proof lzer (previous step) (back to tree)

A spec-lemma