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Proof step 2 from proof shiftright-not-le (previous step)

0 ≤ i ⊦ i ≤ i >> j ↔ i = 0 ∨ j = 0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
0 ≤ i ⊦ i = 0, i ≤ i >> j ↔ i = 0 ∨ j = 0

Rule argument

fma: i = 0

Additional information

Used simplifier rules: shiftright-zero: ⊦ 0 >> j = 0; Used formulas: i = 0