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Proof step 1 from proof abs-2pot (conclusion)

⊦ abs(2 ^ i) = 2 ^ i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ¬ j < 0 → abs(j) = j

Rule argument

spec: int-abs inst: name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [2 ^ i] flags: rewrite

Additional information

Used simplifier rules: pot-zero-01: ⊦ 0 < i → (i ^ j = 0 ↔ j < 0 ∧ 1 < i);%two-lsz: ⊦ ¬ 2 ^ i < 0;%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 ^ i