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Proof step 24 from proof bitsor-bi-ones (previous step)

# bits0 = # bits2 ⊦ # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], bits3 + bits0 = [], bits2int(true ' + bits0 + bits1) = bits2int(true ' + bits1), false ∈ bits3, false ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ (bits + bits0) + bits1 = bits + bits0 + bits1
Proof step 26:
# bits0 = # bits2 ⊦ # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], bits3 + bits0 = [], bits2int((true ' + bits0) + bits1) = bits2int(true ' + bits1), false ∈ bits3, false ∈ bits0

Rule argument

seq: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1 subst: [bits1, bits0, bits] → [bits1, bits0, true '] rotate: T spec: list inst: bitlista name: a

Additional information

Used terms: bits1, bits0, true '