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Proof step 9 from proof sgn-last-04 (previous step)

bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ boolvar1 ↔ boolvar, sgn((boolvar ' + bits0) + boolvar0 ') = sgn(boolvar ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar), ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ bits ≠ [] ∧ (bits.first ↔ ¬ boolvar) → sgn(boolvar ' + bits) = boolvar ' + bits
Proof step 11:
bits0 + boolvar0 ' ≠ [] ∧ ((bits0 + boolvar0 ').first ↔ ¬ boolvar), bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ boolvar1 ↔ boolvar, boolvar ' + bits0 + boolvar0 ' = sgn(boolvar ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar), ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0

Rule argument

name: sgn-base-03 rotate: F subst: [bits, boolvar] → [bits0 + boolvar0 ', boolvar]

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: bits0 + boolvar0 ', boolvar