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Proof step 29 from proof bitsand-first-samesize (previous step)

# bits2 = # bits3, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < #(bits3 + boolvar2 ') → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bitsand(boolvar ' + bits2 + boolvar1 ', boolvar0 ' + bits3 + boolvar2 ') = (boolvar ∧ boolvar0)' + bitsand(bits2, bits3) + (boolvar1 ∧ boolvar2)', bits3 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 31:
1 < #(boolvar ' + bits2 + boolvar1 ') ∧ 1 < #(boolvar0 ' + bits3 + boolvar2 '), # bits2 = # bits3, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < #(bits3 + boolvar2 ') → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bitsand((boolvar ' + bits2 + boolvar1 ').butlast, (boolvar0 ' + bits3 + boolvar2 ').butlast) + ((boolvar ' + bits2 + boolvar1 ').last ∧ (boolvar0 ' + bits3 + boolvar2 ').last) = (boolvar ∧ boolvar0)' + bitsand(bits2, bits3) + (boolvar1 ∧ boolvar2)', bits3 = []

Rule argument

name: and-rec rotate: F subst: [bits0, bits] → [boolvar0 ' + bits3 + boolvar2 ', boolvar ' + bits2 + boolvar1 ']

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;%n < n0 ⊦ (* n) ≠ m0 + (* n0); Used terms: boolvar0 ' + bits3 + boolvar2 ', boolvar ' + bits2 + boolvar1 '