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Proof step 11 from proof cong-mult (previous step)

j0 * j + i = i1 + i0, j ≤ 0 ⊦ j | i1 ↔ j | i - i0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, i0, i2] flags:

Additional information

Used simplifier rules: add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%divides-mult-add: ⊦ i | i * j + j0 ↔ i | j0; from specification int-gcd%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%divides-add: ⊦ i | j → (i | j + j0 ↔ i | j0); from specification int-gcd%div-mult: ⊦ i | i * j; from specification int-gcd% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, i0, i2