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Proof step 6 from proof bitsor-bi-sgn (previous step)

∀ bits, bits1. # bits1 < # bits0 → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits = [], bits0 = [], bits2int(bitsor(sgn(bits), bits0)) = bits2int(bitsor(bits, bits0)), # bits ≠ 1 ∧ # bits0 ≠ 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsor(bits, bits0) = true '
Proof step 8:
¬(bits ≠ [] ∧ bits0 = true '), ∀ bits, bits1. # bits1 < # bits0 → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits = [], bits0 = [], bits2int(bitsor(sgn(bits), bits0)) = bits2int(bitsor(bits, bits0)), # bits ≠ 1 ∧ # bits0 ≠ 1
Proof step 23:
bits ≠ [] ∧ bits0 = true ', ∀ bits, bits1. # bits1 < # bits0 → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits = [], bits0 = [], bits2int(bitsor(sgn(bits), bits0)) = bits2int(true '), # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

name: or-rec-01 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used terms: bits0, bits