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Proof step 2 from proof mod-pred-01 (previous step)

(i - 1) % j + 1 = j, 0 < j, 0 < i, 0 ≤ (i - 1) % j ⊦ i % j = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: mod-succ-02: ⊦ i % j + 1 = j ∧ 0 ≤ i ∧ 0 < j → (i + 1) % j = 0;%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i0