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Proof step 11 from proof ls-pos (previous step)

0 < i, ∀ i1, j. 0 < i1 → (i0 * i1 + i1 < i1 * j ↔ i0 + 1 < j) ⊦ i * i0 < i * j0 ↔ i0 < j0

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: i, j0 + 1 abort: F discard: T computed: F

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;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2 Used terms: i, j0 + 1