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Proof step 29 from proof congmod-inverse (previous step)

i * i0 = i1 * abs(j) + 1, gcd(i, j) = 1, 0 < i0, 0 ≤ i, 1 < abs(j), 0 ≤ i1 ⊦ j = 0, j | i1 * abs(j) + 1 - 1, ∃ i0. ≡mod(i * i0, 1, j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ i + j - j = i

Rule argument

spec: int-abs inst: name: sub-same-01 seq: ⊦ i + j - j = i subst: [j, i] → [1, i1 * abs(j)] flags: rewrite

Additional information

Used simplifier rules: div-abs: ⊦ i | abs(i) * j; from specification int-gcd%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 1, i1 * abs(j)