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Proof step 6 from proof i2n (previous step)

0 ≤ j, i < 0 ⊦ i→n(i) * i→n(j) = i→n(i * j)

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: i2np seq: 0 ≤ i, 0 ≤ j ⊦ i→n(i) * i→n(j) = i→n(i * j) subst: [i, j] → [~ i, j] flags: current, []

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%minus-left: ⊦ ~ i * j = ~(i * 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%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; 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%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: ~ i, j