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Proof step 10 from proof int2bits-lslen-06 (previous step)

0 < i, # int2bits(i) ≤ m, i→n(n→i(m) - 1) + 1 < # int2bits(i), i→n(n→i(m) - 1) + 1 < m ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ 1 < i → i→n(i - 1) = i→n(i) -1
Proof step 12:
1 < n→i(m), 0 < i, # int2bits(i) ≤ m, i→n(n→i(m)) -1 + 1 < # int2bits(i), i→n(n→i(m)) -1 + 1 < m ⊦

Rule argument

seq: ⊦ 1 < i → i→n(i - 1) = i→n(i) -1 subst: [i] → [n→i(m)] rotate: F spec: intnat inst: name: rax-01

Additional information

Used simplifier rules: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m < n → n ≠ 0;% ⊦ m < n ∧ n = n0 → m < n0; (forward) Used terms: n→i(m)