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Proof step 29 from proof byte2bits-val (previous step)

0 ≤ n→i(# int2bits(b2i(by))) ∧ 0 ≤ n→i(n), 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ n = 0, 2 ^ n→i(# int2bits(b2i(by)) + n) = 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ m + n = n + m
Proof step 31:
0 ≤ n→i(# int2bits(b2i(by))) ∧ 0 ≤ n→i(n), 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ n = 0, 2 ^ n→i(n + # int2bits(b2i(by))) = 256

Rule argument

seq: ⊦ m + n = n + m subst: [n, m] → [n, # int2bits(b2i(by))] rotate: F spec: nat-basic2 inst: name: c

Additional information

Used simplifier rules: ⊦ m + n = n + m; Used terms: n, # int2bits(b2i(by))