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Proof step 4 from proof mod-sub-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: mod-add-01 seq: ⊦ 0 < i ∧ j < 0 ∧ i ≤ abs(j) → (i + j) % i = j % i subst: [j, i] → [~ j, i] flags: <all paths>

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);% ⊦ abs(~ i) = abs(i); (s)% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ ~ i = ~ j ↔ i = j; (s)% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i - 0 = i; (s) Used terms: ~ j, i