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Proof step 5 from proof int2bits-less-2pot (previous step)

0 ≤ j, 2 ^ (n→i(# int2bits(j)) - 1) ≤ j, 2 + i→n(n→i(# int2bits(j)) - 1) ≤ # uint2bits(j) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 1 < i → i→n(i - 1) = i→n(i) -1
Proof step 7:
¬ 1 < n→i(# int2bits(j)), 0 ≤ j, 2 ^ (n→i(# int2bits(j)) - 1) ≤ j, 2 + i→n(n→i(# int2bits(j)) - 1) ≤ # uint2bits(j) ⊦
Proof step 8:
1 < n→i(# int2bits(j)), 0 ≤ j, 2 ^ (n→i(# int2bits(j)) - 1) ≤ j, 2 + i→n(n→i(# int2bits(j))) -1 ≤ # uint2bits(j) ⊦

Rule argument

seq: ⊦ 1 < i → i→n(i - 1) = i→n(i) -1 subst: [i] → [n→i(# int2bits(j))] rotate: F spec: intnat inst: name: rax-01

Additional information

Used terms: n→i(# int2bits(j))