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Proof step 3 from proof numls-07 (previous step)

⊦ i1 * j < j ↔ 0 < (1 - i1) * j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i → (j * i < i ↔ j < 1)
Proof step 5:
⊦ i1 * j < j ↔ 0 < (1 - i1) * j, 0 < j
Proof step 14:
0 < j ⊦ i1 < 1 ↔ 0 < (1 - i1) * j

Rule argument

spec: <toplevel> name: ls-same-04 seq: ⊦ 0 < i → (j * i < i ↔ j < 1) subst: [i, j] → [j, i1] flags: <all paths>

Additional information

Used terms: j, i1