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Proof step 4 from proof congmod-mod-abs (previous step)

⊦ j | i - i0 ↔ ≡mod(i, i0, abs(j))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
i = i1 + i0 ⊦ j | i1 ↔ ≡mod(i, i0, abs(j))

Rule argument

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

Additional information

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