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Proof step 21 from proof int2bits-leneq-neg (previous step)

n→i(# int2bits(j)) = i0 + 2, j < ~1, j < 0, # int2bits(j) ≤ i→n(i0) + 1, 2 < # int2bits(j), i0 < ~2, i0 < 0 ⊦ i0 = 0, # int2bits(j) = 2, # int2bits(j) = 3, # int2bits(j) = 1, ~1 = i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
i < 0 ⊦ n→i(n) ≠ (* i)

Rule argument

spec: intnat inst: name: num-n2i-eq-01 seq: i < 0 ⊦ n→i(n) ≠ (* i) subst: [i, n] → [i0 + 2, # int2bits(j)] flags: rewrite

Additional information

Used simplifier rules: numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i0 + 2, # int2bits(j)