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

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

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-01: ⊦ i % j + 1 ≠ j ∧ 0 ≤ i ∧ 0 < j → (i + 1) % j = i % j + 1;%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%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i0