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Proof step 16 from proof mod-pred-02 (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); 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%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs