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Proof step 1 from proof exinpot (conclusion)
0 < j, 1 < i ⊦ ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j ∧ j < i ^ (i0 + 1)
System applied rule normalize
And obtained the following new goals
Proof step 2
:
1 < i, 0 < j ⊦ ∃ i0. 0 ≤ i0 ∧ i ^ i0 ≤ j ∧ j < i ^ (i0 + 1)
Rule argument
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