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Proof step 42 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(bitsor(boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(boolvar ' + bits0, bits3) + (boolvar ∨ boolvar0)'), bits3 = [], (¬ boolvar) ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 43:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsor(bits, bits0) = true '
Proof step 44:
¬(bits3 + boolvar0 ' ≠ [] ∧ boolvar ' = true '), ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(bitsor(boolvar ', bits3 + boolvar0 ')) = bits2int(bitsor(boolvar ' + bits0, bits3) + (boolvar ∨ boolvar0)'), bits3 = [], (¬ boolvar) ∈ bits0
Proof step 47:
bits3 + boolvar0 ' ≠ [] ∧ boolvar ' = true ', ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits2int(true ') = bits2int(bitsor(boolvar ' + bits0, bits3) + (boolvar ∨ boolvar0)'), bits3 = [], (¬ boolvar) ∈ bits0

Rule argument

name: or-rec-03 rotate: F subst: [bits0, bits] → [bits3 + boolvar0 ', boolvar ']

Additional information

Used terms: bits3 + boolvar0 ', boolvar '