Proof step 1 from proof two-mod (conclusion) (back to tree)

⊦ 1 < i → 2 ^ i % 2 = 0

System applied rule normalize

And obtained the following new goals

Proof step 2:
2 ^ i % 2 ≠ 0, 1 < i ⊦

Rule argument

Additional information

Proof step 2 from proof two-mod (previous step) (back to tree)

2 ^ i % 2 ≠ 0, 1 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 ^ i % 2 ≠ 0, 1 < i, 2 ≤ i ⊦

Rule argument

Additional information

Used simplifier rules: numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs

Proof step 3 from proof two-mod (previous step) (back to tree)

2 ^ i % 2 ≠ 0, 1 < i, 2 ≤ i ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → j ^ i = j ^ i -1 * j

Rule argument

spec: <toplevel>, name:Pot-pos, seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j, subst:[i, j] → [i, 2], <all paths>

Additional information

Used simplifier rules: ⊦ (i * 2) % 2 = 0; (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult ⊦ i < j ↔ ¬ j ≤ i; (ss) forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs Used terms: i, 2

Proof step 4 from proof two-mod (previous step) (back to tree)

⊦ 0 < i → j ^ i = j ^ i -1 * j

Lemma Pot-pos

Proof Tree for two-mod (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.