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Proof step 10 from proof int2bytes-byte (previous step)

8 < # int2bits(i), i < 128, ~128 ≤ i ⊦ bits2bytes(firstn(# int2bits(i) - 8, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# int2bits(i), 8, n] flags:

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # int2bits(i), 8, n