Proof step 1 from proof lemon2pot-01 (conclusion) (back to tree)

⊦ j ≤ 2 ^ n→i(m * n - 1) ∧ n ≤ n0 ∧ m ≠ 0 ∧ n ≠ 0 → j ≤ 2 ^ n→i(m * n0 - 1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
m ≠ 0, n ≠ 0, j ≤ 2 ^ n→i(m * n - 1), n ≤ n0, ¬ j ≤ 2 ^ n→i(m * n0 - 1) ⊦

Rule argument

Additional information

Proof step 2 from proof lemon2pot-01 (previous step) (back to tree)

m ≠ 0, n ≠ 0, j ≤ 2 ^ n→i(m * n - 1), n ≤ n0, ¬ j ≤ 2 ^ n→i(m * n0 - 1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ j ≤ 2 ^ i ∧ i ≤ i0 ∧ 0 ≤ i → j ≤ 2 ^ i0
Proof step 4:
m ≠ 0, n ≠ 0, n ≤ n0, j ≤ 2 ^ n→i(m * n - 1), ¬ j ≤ 2 ^ n→i(m * n0 - 1) ⊦ j ≤ 2 ^ n→i(m * n - 1) ∧ n→i(m * n - 1) ≤ n→i(m * n0 - 1) ∧ 0 ≤ n→i(m * n - 1)

Rule argument

spec: <toplevel>, name:lemon2pot, seq: ⊦ j ≤ 2 ^ i ∧ i ≤ i0 ∧ 0 ≤ i → j ≤ 2 ^ i0, subst:[j, i, i0] → [j, n→i(m * n - 1), n→i(m * n0 - 1)], <all paths>

Additional information

Used terms: j, n→i(m * n - 1), n→i(m * n0 - 1)

Proof step 3 from proof lemon2pot-01 (previous step) (back to tree)

⊦ j ≤ 2 ^ i ∧ i ≤ i0 ∧ 0 ≤ i → j ≤ 2 ^ i0

Lemma lemon2pot

Proof step 4 from proof lemon2pot-01 (previous step) (back to tree)

m ≠ 0, n ≠ 0, n ≤ n0, j ≤ 2 ^ n→i(m * n - 1), ¬ j ≤ 2 ^ n→i(m * n0 - 1) ⊦ j ≤ 2 ^ n→i(m * n - 1) ∧ n→i(m * n - 1) ≤ n→i(m * n0 - 1) ∧ 0 ≤ n→i(m * n - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m ≠ 0, n ≠ 0, 2 ^ n→i(m * n0 - 1) < 2 ^ n→i(m * n - 1), (m * n0 - 1) + 1 < m * n, 2 ^ n→i(m * n0 - 1) < j, n ≤ n0, j ≤ 2 ^ n→i(m * n - 1) ⊦

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws) forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ 0 ≤ n→i(n); (s) n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n; (s) ⊦ i ≤ j ↔ ¬ j < i; (ss)

Proof step 5 from proof lemon2pot-01 (previous step) (back to tree)

m ≠ 0, n ≠ 0, 2 ^ n→i(m * n0 - 1) < 2 ^ n→i(m * n - 1), (m * n0 - 1) + 1 < m * n, 2 ^ n→i(m * n0 - 1) < j, n ≤ n0, j ≤ 2 ^ n→i(m * n - 1) ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [m * n0, m0]

Additional information

Used simplifier rules: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws) forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n → n ≠ 0; (s) ⊦ ¬ n < n; (s) ⊦ m ≤ 0 ↔ m = 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m * n0, m0

Proof Tree for lemon2pot-01 (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.