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Proof step 17 from proof congmod-inverse-smallest (previous step)

gcd(i, j) = 1, 0 ≤ i0, 0 < i, i0 |%| j < j, i0 |%| j < i0, 0 ≤ i0 |%| j, 0 < j, j ≤ i0, 0 < i0, j | i * i0 - 1 ⊦ i = 0, j = 0, (i * i0 |%| j) |%| j = 1 |%| j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
i * i0 = i1 + 1, gcd(i, j) = 1, 0 ≤ i0, 0 < i, i0 |%| j < j, i0 |%| j < i0, 0 ≤ i0 |%| j, 0 < j, j ≤ i0, 0 < i0, j | i1 ⊦ i = 0, j = 0, (i * i0 |%| j) |%| j = 1 |%| j

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i * i0, 1, i1] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i * i0, 1, i1