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Proof step 9 from proof int2bits-leneq-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j)
Proof step 11:
2 ≤ n→i(# int2bits(j)) ∧ 0 ≤ 2, # int2bits(j) ≤ (i→n(n→i(# int2bits(j))) - i→n(2)) + 1, 0 < j, 2 < # int2bits(j) ⊦ j = 0, # int2bits(j) = 3, # int2bits(j) = 2

Rule argument

spec: intnat inst: name: sub seq: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j) subst: [j, i] → [2, n→i(# int2bits(j))] flags: rewrite

Additional information

Used simplifier rules: num-n2i-ls-06: 0 ≤ i ⊦ (* i) ≤ n→i(n) ↔ i→n(i) ≤ n; from specification intnat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: 2, n→i(# int2bits(j))