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Proof step 13 from proof square-nls (previous step)

j * (j + 1) = i0 * 2 + j0, (i0 + 1) * j < (i0 + 1) * 2, i0 < j, j0 < 2, 0 < j, 0 ≤ i0, 0 ≤ j0 ⊦ j = 0, ~1 = i0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: ls-pos seq: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0) subst: [j0, j, i] → [2, j, i0 + 1] flags: rewrite

Additional information

Used simplifier rules: 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%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%two-ls: ⊦ 0 < j → (j < 2 ↔ j = 1); from specification int-basic1%one-01: ⊦ i = 1 → i * j = 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%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%bet-01: ⊦ 0 ≤ i → (i < 1 ↔ i = 0); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1 Used terms: 2, j, i0 + 1