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Proof step 21 from proof prime-def-mod (previous step)

1 < i, 0 < i, 1 < i0, 0 < i0, i0 < i, 2 ≤ i * i0, 0 ≤ i0, ∀ i1. 1 < i1 ∧ i1 * i1 ≤ i * i0 → (i * i0) % i1 ≠ 0 ⊦ i * i0 = 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: i0 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i * 0 = 0; (s)%j ≠ 0 ⊦ (i * j) % j = 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: i0