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Proof step 45 from proof bitsor-sgn-in (previous step)

⊦ bits = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits, false ' + bits1), true ∈ bits

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 46:
⊦ ¬ true ∈ bits ∧ ¬ bits0.first ∧ # bits ≤ # bits0 ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) = bits0
Proof step 47:
¬(¬ true ∈ bits ∧ ¬ (false ' + bits1).first ∧ # bits ≤ #(false ' + bits1) ∧ bits ≠ [] ∧ false ' + bits1 ≠ []) ⊦ bits = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits, false ' + bits1), true ∈ bits

Rule argument

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

Additional information

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