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Proof step 16 from proof and-false (previous step)

bitsand(true ' + bits3, true ' + bits2) + (boolvar ∧ boolvar0)' = false ' + bits1, ∀ bits1, bits, bits0. # bits0 < #(bits3 + boolvar ') → bitsand(true ' + bits0, true ' + bits) = false ' + bits1 → false ⊦ boolvar ∧ boolvar0 ∧ ¬ true ∈ bits1, ¬ false ∈ bitsand(true ' + bits3, true ' + bits2) ∧ boolvar ∧ boolvar0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
bits ≠ [] ⊦ bits0 + boolvar ' = bits ↔ bits0 = bits.butlast ∧ (boolvar ↔ bits.last)
Proof step 18:
false ' + bits1 ≠ [], bitsand(true ' + bits3, true ' + bits2) = (false ' + bits1).butlast ∧ (boolvar ∧ boolvar0 ↔ (false ' + bits1).last), ∀ bits1, bits, bits0. # bits0 < #(bits3 + boolvar ') → bitsand(true ' + bits0, true ' + bits) = false ' + bits1 → false ⊦ boolvar ∧ boolvar0 ∧ ¬ true ∈ bits1, ¬ false ∈ bitsand(true ' + bits3, true ' + bits2) ∧ boolvar ∧ boolvar0

Rule argument

seq: bits ≠ [] ⊦ bits0 + boolvar ' = bits ↔ bits0 = bits.butlast ∧ (boolvar ↔ bits.last) subst: [bits, boolvar, bits0] → [false ' + bits1, boolvar ∧ boolvar0, bitsand(true ' + bits3, true ' + bits2)] rotate: F spec: list-last inst: bitlista name: splitlasteq

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: false ' + bits1, boolvar ∧ boolvar0, bitsand(true ' + bits3, true ' + bits2)