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

i < j, 0 ≤ i ⊦ 2 ^ i % 2 ^ j = 2 ^ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i; (s)%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs