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Proof step 63 from proof bitsor-bi-sgn (previous step)

∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bits3 + true ') = bits2int(bitsor(false ' + bits0, bits3) + true '), bits3 = [], true ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 64:
⊦ ¬ true ∈ bits ∧ bits ≠ [] ∧ bits0 ≠ [] → bits2int(bitsor(bits, bits0) + bits1) = bits2int(bits0 + bits1)

Rule argument

name: bitsor-bi-zeroes rotate: F subst: [bits1, bits0, bits] → [true ', bits3, false ' + bits0]

Additional information

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