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Proof step 6 from proof bitsand-zeros (previous step)

# bits ≤ 1 + # bits1 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ bitsand(bits, boolvar ' + bits1) = bits, bits = [], true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 8:
¬(1 < # bits ∧ 1 < #(boolvar ' + bits1)), # bits ≤ 1 + # bits1 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ bitsand(bits, boolvar ' + bits1) = bits, bits = [], true ∈ bits
Proof step 13:
1 < # bits ∧ 1 < #(boolvar ' + bits1), # bits ≤ 1 + # bits1 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ bitsand(bits.butlast, (boolvar ' + bits1).butlast) + (bits.last ∧ (boolvar ' + bits1).last) = bits, bits = [], true ∈ bits

Rule argument

name: and-rec rotate: F subst: [bits0, bits] → [boolvar ' + bits1, bits]

Additional information

Used terms: boolvar ' + bits1, bits