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Proof step 1 from proof shiftright-not-le (conclusion)
⊦ 0 ≤ i → (i ≤ i >> j ↔ i = 0 ∨ j = 0)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
0 ≤ i ⊦ i ≤ i >> j ↔ i = 0 ∨ j = 0
Rule argument
Additional information
Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)