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Proof step 17 from proof div-mult-two-nls-01 (previous step)

(2 * i0 + 1) * j + (2 * j0 - j) = i1 * j + j1, j1 < 2 * j0, j ≤ 2 * j0, 0 ≤ i1, 0 ≤ j1, 0 ≤ i0, 0 ≤ j0, 1 + 2 * i0 < i1, j0 < j, j1 < j, 1 < j, 0 < j, 0 < j0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: mods-eq seq: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j0 < j ∧ j1 < j → (i * j + j0 = i0 * j + j1 ↔ i = i0 ∧ j0 = j1) subst: [j1, i0, j0, j, i] → [j1, i1, 2 * j0 - j, j, 2 * i0 + 1] flags: rewrite

Additional information

Used simplifier rules: zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * 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%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); from specification int-mult%numls-01: ⊦ (* i2) * i < (* i1) * i + j0 ↔ (i2 - i1) * i < j0; from specification int-mult%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%eq-succ-01: ⊦ (i + i0) * j + j0 = (i + i1) * j + j1 ↔ i0 * j + j0 = i1 * j + j1; from specification int-mult%dist-rmerge: ⊦ i * j + i * j0 = i * (j + j0); from specification int-mult Used terms: j1, i1, 2 * j0 - j, j, 2 * i0 + 1