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Proof step 15 from proof bitsand-first-samesize (previous step)

# bits1 = 1, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < 1 → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bits1 = [], bitsand(boolvar ' + bits1, boolvar0 ' + false ') = (boolvar ∧ boolvar0)' + false '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 17:
1 < #(boolvar ' + bits1) ∧ 1 < #(boolvar0 ' + false '), # bits1 = 1, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < 1 → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bits1 = [], bitsand((boolvar ' + bits1).butlast, (boolvar0 ' + false ').butlast) + ((boolvar ' + bits1).last ∧ (boolvar0 ' + false ').last) = (boolvar ∧ boolvar0)' + false '

Rule argument

name: and-rec rotate: F subst: [bits0, bits] → [boolvar0 ' + false ', boolvar ' + bits1]

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 + 1 ≠ 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = []; Used terms: boolvar0 ' + false ', boolvar ' + bits1