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Proof step 5 from proof mod-two-ls (previous step)

0 < j, i < j, 0 ≤ i ⊦ 2 ^ i % 2 ^ j = 2 ^ i, 0 < 2 ^ j ∧ 0 ≤ 2 ^ i ∧ 2 ^ i < 2 ^ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, i < j, 0 ≤ i, 0 ≤ 2 ^ i ⊦ 2 ^ i % 2 ^ j = 2 ^ i, 2 ^ i < 2 ^ j ∧ 0 < 2 ^ j

Rule argument

Additional information

Used simplifier rules: two-lsz: ⊦ ¬ 2 ^ i < 0;% ⊦ i ≤ j ↔ ¬ j < i; (ss)