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Proof step 10 from proof divides-prime (previous step)

2 ≤ i * i0, 0 ≤ i * i0, 0 < i * i0, ∀ i1. 1 < i1 ∧ i1 < i * i0 → ¬ i1 | i * i0 ⊦ abs(i) = i * i0, abs(i) = 1

User applied rule all left

And obtained the following new goals

Proof step 11:
2 ≤ i * i0, 0 ≤ i * i0, 0 < i * i0, 1 < i0 ∧ i0 < i * i0 → ¬ i0 | i * i0, ∀ i1. 1 < i1 ∧ i1 < i * i0 → ¬ i1 | i * i0 ⊦ abs(i) = i * i0, abs(i) = 1

Rule argument

pos: left 4 terms: i0 abort: F discard: F computed: F

Additional information

Used terms: i0