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Proof step 15 from proof fac-le (previous step)

0 ≤ i1, 0 < i1, (i1 + 1) * fac(i1) ≤ 1, ∀ i0, j. i→n(i0) < i→n(i1 + 1) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: le-one seq: ⊦ 1 < j → (i * j ≤ 1 ↔ i ≤ 0) subst: [j, i] → [fac(i1), i1 + 1] flags: rewrite

Additional information

Used simplifier rules: fac-le-one: ⊦ fac(i) ≤ 1 ↔ i ≤ 1;%fac-one: ⊦ fac(1) = 1;%numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs%mone-05: ⊦ i ≤ ~1 ↔ i < 0; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: fac(i1), i1 + 1