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Proof step 2 from proof and-rec-08 (previous step)

⊦ bitsand(bits + boolvar ' + boolvar0 ', boolvar1 ' + boolvar2 ') = bitsand(bits + boolvar ', boolvar1 ') + (boolvar0 ∧ boolvar2)'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)

Rule argument

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

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + boolvar ' + bits0).last ↔ (boolvar ' + bits0).last;% ⊦ (bits + boolvar ').butlast = bits;%⊦ (bits + boolvar ' + bits0).butlast = bits + (boolvar ' + bits0).butlast;% ⊦ #(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;% ⊦ # bits = 0 ↔ bits = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;%n < n0 ⊦ (* n) ≠ m0 + (* n0); Used terms: boolvar1 ' + boolvar2 ', bits + boolvar ' + boolvar0 '