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Proof step 15 from proof bitsor-app (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ bits ≠ [] ∧ bits0 = true ' → bitsor(bits, bits0) = true '
Proof step 17:
¬(boolvar ' ≠ [] ∧ boolvar0 ' = true '), ∀ bits1, bits, bits2, bits3. # bits3 < #([] + boolvar ') → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits1 = [], bits = [], (boolvar ∨ boolvar0)' = bitsor(boolvar ', boolvar0 ')

Rule argument

name: or-rec-01 rotate: F subst: [bits0, bits] → [boolvar0 ', boolvar ']

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: boolvar0 ', boolvar '