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

j ≤ i, 0 ≤ j ⊦ 2 ^ i % 2 ^ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≤ j, j ≤ i, 0 ≤ i ⊦ 2 ^ i % 2 ^ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~1 < i ↔ 0 ≤ i; (s)%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs