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Proof step 5 from proof bits2int-sign (previous step)

⊦ bits0 = [], bits2int(bits0 + boolvar ' + bits1) = bits2int(boolvar ' + bits1), (¬ boolvar) ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ¬ bits.first ∧ bits ≠ [] → bits2int(bits) = bits2uint(bits.rest)
Proof step 7:
¬(¬ (boolvar ' + bits1).first ∧ boolvar ' + bits1 ≠ []) ⊦ bits0 = [], bits2int(bits0 + boolvar ' + bits1) = bits2int(boolvar ' + bits1), (¬ boolvar) ∈ bits0
Proof step 17:
¬ (boolvar ' + bits1).first ∧ boolvar ' + bits1 ≠ [] ⊦ bits0 = [], bits2int(bits0 + boolvar ' + bits1) = bits2uint((boolvar ' + bits1).rest), (¬ boolvar) ∈ bits0

Rule argument

name: bits2int-pos-01 rotate: F subst: [bits] → [boolvar ' + bits1]

Additional information

Used terms: boolvar ' + bits1