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Proof step 7 from proof square-interval (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: square-ls-02 seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (i * (i + 1) < j * (j + 1) ↔ i < j) subst: [j, i] → [i + 1, j] flags: <all paths>

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0;%dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%numls-04: ⊦ i + j < (* i1) * i + j0 ↔ j < (i1 - 1) * i + j0;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j < i + i0 ↔ j < i0; (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ i < 0 ∧ 0 ≤ j → ¬ j < i; (s) Used terms: i + 1, j