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Proof step 18 from proof bitsand-fillsgn (previous step)

# bits = 1, ∀ n, bits, bits1. # bits1 < # bits0 → 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 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsand(bits, bits0) = bits0
Proof step 20:
¬(bits0 ≠ [] ∧ bits = true '), # bits = 1, ∀ n, bits, bits1. # bits1 < # bits0 → 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 = []
Proof step 38:
bits0 ≠ [] ∧ bits = true ', # bits = 1, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bits0), bits0 = [], bits = []

Rule argument

name: and-rec-03 rotate: F subst: [bits0, bits] → [bits0, bits]

Additional information

Used terms: bits0, bits