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Proof step 9 from proof mod-ls-same-02 (previous step)

i < 0 ⊦ i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ abs(i) < j → i % j = i
Proof step 11:
i < 0 ⊦ i < i % j ↔ abs(j) = 1 ∨ ¬ 0 < abs(j) + i, j = 0, abs(i) < j

Rule argument

spec: <toplevel> name: mod-zero-05 seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ i < i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ ¬ abs(i) < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j, i