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Proof step 34 from proof bitsor-bi-zeroes (previous step)

# bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int((bits3 + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], true ∈ bits3, true ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 35:
⊦ ¬ bits.first ∧ bits ≠ [] → bits2int(bits) = bits2uint(bits)
Proof step 36:
¬(¬ ((bits3 + bitsor(bits0, bits2)) + bits1).first ∧ (bits3 + bitsor(bits0, bits2)) + bits1 ≠ []), # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int((bits3 + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], true ∈ bits3, true ∈ bits0
Proof step 38:
¬ ((bits3 + bitsor(bits0, bits2)) + bits1).first ∧ (bits3 + bitsor(bits0, bits2)) + bits1 ≠ [], # bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2uint((bits3 + bitsor(bits0, bits2)) + bits1) = bits2uint(bits2 + bits1), # bits3 = 1, bits2 = [], bits0 = [], bits3 = [], true ∈ bits3, true ∈ bits0

Rule argument

seq: ⊦ ¬ bits.first ∧ bits ≠ [] → bits2int(bits) = bits2uint(bits) subst: [bits] → [(bits3 + bitsor(bits0, bits2)) + bits1] rotate: F spec: int2bits inst: name: bits2int-pos-02

Additional information

Used terms: (bits3 + bitsor(bits0, bits2)) + bits1