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Proof step 25 from proof eq-one (previous step)

i * j = 1, 0 < i, 0 ≤ j, j ≤ 1, 0 ≤ i ⊦ i = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ i ≤ j → (j ≤ i ↔ i = j)

Rule argument

spec: int-abs name: anti seq: ⊦ i ≤ j → (j ≤ i ↔ i = j) subst: [i, j] → [1, j] flags: <all paths>

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s) Used terms: 1, j