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Proof step 41 from proof and-sgn (previous step)

∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = [], sgn(bitsand(boolvar0 ', bits2 + boolvar ')) = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits0 + boolvar0 ' = bits0.first ', (¬ boolvar0) ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsand(bits, bits0) = bits0
Proof step 43:
¬(bits2 + boolvar ' ≠ [] ∧ boolvar0 ' = true '), ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = [], sgn(bitsand(boolvar0 ', bits2 + boolvar ')) = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits0 + boolvar0 ' = bits0.first ', (¬ boolvar0) ∈ bits0
Proof step 49:
bits2 + boolvar ' ≠ [] ∧ boolvar0 ' = true ', ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = [], sgn(bits2 + boolvar ') = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits0 + boolvar0 ' = bits0.first ', (¬ boolvar0) ∈ bits0

Rule argument

name: and-rec-03 rotate: F subst: [bits0, bits] → [bits2 + boolvar ', boolvar0 ']

Additional information

Used terms: bits2 + boolvar ', boolvar0 '