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Proof step 2 from proof shiftright-not-ls-01 (previous step)

i < i >> j, 0 ≤ i, 0 < i >> j ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ i >> j ≤ abs(i)

Rule argument

name: shiftright-is-le subst: [i, j] → [i, j]

Additional information

Used simplifier rules: ⊦ 0 ≤ i → abs(i) = i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: i, j