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Proof step 5 from proof gcd-mod-one-01 (previous step)

(i * i0) % j = 1, gcd(i, j) = 1, 0 < j, 1 < j, 0 < i, 1 < i ⊦ (i * i0) % j = 1 ∧ 0 < i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(i * i0) % j = 1, gcd(i, j) = 1, 0 < j, 1 < j, 0 < i, 1 < i, i0 ≤ 0, i0 < j, i0 < i ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs