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Proof step 3 from proof gcd-mod-one-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(i * i0) % j = 1, gcd(i, j) = 1, 0 < j, 1 < j, 0 < i, 1 < i ⊦ ∃ i0. (i * i0) % j = 1 ∧ 0 < i0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1