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Proof step 48 from proof ex-modeq-half (previous step)

odd(j), 0 < j, j / 2 < i0 % j, 0 ≤ i0 % j, i0 % j + j / 2 < j, 0 < i0 % j ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-min inst: name: odd-def seq: ⊦ odd(i) ↔ (∃ j. i = j + j + 1) subst: [i] → [j] flags: rewrite

Additional information

Used simplifier rules: add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%div-two: ⊦ 0 ≤ i → (2 * i + 1) / 2 = i; from specification int-div%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j