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Proof step 18 from proof bitsand-app-true (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 20:
1 < #(bits2 + boolvar0 ') ∧ 1 < #(true ' + bits3 + boolvar '), # bits3 < # bits2, ∀ bits, bits0, bits1. # bits1 < #(bits3 + boolvar ') → # bits1 < # bits0 → bitsand(bits + bits0, true ' + bits1) = bits + bitsand(bits0, true ' + bits1) ∨ bits0 = [] ⊦ bits2 = [], bitsand(bits + bits2, true ' + bits3) + (boolvar0 ∧ boolvar)' = bits + (bitsand((bits2 + boolvar0 ').butlast, (true ' + bits3 + boolvar ').butlast) + ((bits2 + boolvar0 ').last ∧ (true ' + bits3 + boolvar ').last))

Rule argument

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

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 = [];% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;%n < n0 ⊦ (* n) ≠ m0 + (* n0); Used terms: true ' + bits3 + boolvar ', bits2 + boolvar0 '