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Proof step 3 from proof two-mod-01 (previous step)

1 < i, 2 ≤ i ⊦ (2 ^ i - 1) % 2 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 5:
1 < i, 0 < i, 2 ≤ i ⊦ (2 ^ i -1 * 2 - 1) % 2 = 1

Rule argument

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

Additional information

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