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Proof step 2 from proof mod-two-ls (previous step)

i < j, 0 ≤ i, 0 < j ⊦ 2 ^ i % 2 ^ j = 2 ^ i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
0 < j, 0 ≤ i, i < j ⊦ i % j = i

Rule argument

spec: int-div inst: name: mod-zero-03 seq: 0 < j, 0 ≤ i, i < j ⊦ i % j = i subst: [j, i] → [2 ^ j, 2 ^ i] flags: rewrite

Additional information

Used simplifier rules: pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%ls-pos-2pot: ⊦ 0 ≤ j → (2 ^ j < 2 ^ j0 ↔ j < j0);%two-lsz: ⊦ ¬ 2 ^ i < 0;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: 2 ^ j, 2 ^ i