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Proof step 39 from proof and-insgn (previous step)

∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(boolvar1 ', bits2 + boolvar0 ') ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', (¬ boolvar1) ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 40:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsand(bits, bits0) = bits0
Proof step 41:
¬(bits2 + boolvar0 ' ≠ [] ∧ boolvar1 ' = true '), ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(boolvar1 ', bits2 + boolvar0 ') ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', (¬ boolvar1) ∈ bits0
Proof step 44:
bits2 + boolvar0 ' ≠ [] ∧ boolvar1 ' = true ', ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bits2 + boolvar0 ' ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', (¬ boolvar1) ∈ bits0

Rule argument

name: and-rec-03 rotate: F subst: [bits0, bits] → [bits2 + boolvar0 ', boolvar1 ']

Additional information

Used terms: bits2 + boolvar0 ', boolvar1 '