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Proof step 6 from proof bits2int-2pot (previous step)

⊦ 2 ^ n→i(# bits) + bits2int(sgn(bytes2bits(bits2bytes(true ' + bits)))) = 0, true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: bits2bits-sgn rotate: F subst: [bits] → [true ' + bits]

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ ¬ boolvar ∈ bits → sgn(boolvar ' + bits) = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ true ∈ bits → bits2int(true ' + bits) = ~(2 ^ n→i(# bits));% ⊦ ~ i + i = 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: true ' + bits