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Proof step 18 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) ⊦ # int2bits(j) = 2, # int2bits(j) = 3, # int2bits(j) = 1, i→n(i0) = 1, i0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
0 ≤ i ⊦ n→i(n) = (* i) ↔ n = i→n(i)
Proof step 20:
¬ 0 ≤ i0 + 2, n→i(# int2bits(j)) = i0 + 2, j < ~1, j < 0, # int2bits(j) ≤ i→n(i0) + 1, 2 < # int2bits(j) ⊦ # int2bits(j) = 2, # int2bits(j) = 3, # int2bits(j) = 1, i→n(i0) = 1, i0 = 0
Proof step 23:
0 ≤ i0 + 2, # int2bits(j) = i→n(i0 + 2), j < ~1, j < 0, # int2bits(j) ≤ i→n(i0) + 1, 2 < # int2bits(j) ⊦ # int2bits(j) = 2, # int2bits(j) = 3, # int2bits(j) = 1, i→n(i0) = 1, i0 = 0

Rule argument

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

Additional information

Used terms: i0 + 2, # int2bits(j)