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Proof step 10 from proof bits2int-lasteq (previous step)

⊦ bits0 = [], bits2int((boolvar0 ' + bits1) + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2int(boolvar0 ' + bits1) = bits2int(bits0), boolvar ∧ ¬ boolvar0, boolvar0 ∧ boolvar

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ ¬ bits.first ∧ bits ≠ [] → bits2int(bits) = bits2uint(bits.rest)
Proof step 12:
¬(¬ (boolvar0 ' + bits1).first ∧ boolvar0 ' + bits1 ≠ []) ⊦ bits0 = [], bits2int((boolvar0 ' + bits1) + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2int(boolvar0 ' + bits1) = bits2int(bits0), boolvar ∧ ¬ boolvar0, boolvar0 ∧ boolvar
Proof step 19:
¬ (boolvar0 ' + bits1).first ∧ boolvar0 ' + bits1 ≠ [] ⊦ bits0 = [], bits2int((boolvar0 ' + bits1) + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2uint((boolvar0 ' + bits1).rest) = bits2int(bits0), boolvar ∧ ¬ boolvar0, boolvar0 ∧ boolvar

Rule argument

name: bits2int-pos-01 rotate: F subst: [bits] → [boolvar0 ' + bits1]

Additional information

Used terms: boolvar0 ' + bits1