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Proof step 9 from proof bitsand-fillsgn (previous step)

# bits0 = 1, ∀ n, bits, bits1. # bits1 < 1 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = [], # bits = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsand(bits, bits0) = bits
Proof step 11:
¬(bits ≠ [] ∧ bits0 = true '), # bits0 = 1, ∀ n, bits, bits1. # bits1 < 1 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = [], # bits = 1
Proof step 14:
bits ≠ [] ∧ bits0 = true ', # bits0 = 1, ∀ n, bits, bits1. # bits1 < 1 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bits), bits0 = [], bits = [], # bits = 1

Rule argument

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

Additional information

Used terms: bits0, bits