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Proof step 5 from proof int2bits-less-2pot-02 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

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

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [n→i(# int2bits(j)), 1, i] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: n→i(# int2bits(j)), 1, i