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Proof step 19 from proof sgn-last-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ bits ≠ [] ∧ (bits.first ↔ ¬ boolvar) → sgn(boolvar ' + bits) = boolvar ' + bits
Proof step 21:
¬(bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar1)), bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(boolvar1 ' + bits) = sgn(boolvar1 ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1), ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0
Proof step 39:
bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar1), bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(boolvar1 ' + bits) = (boolvar1 ' + bits0) + boolvar0 ', ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1), ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0

Rule argument

name: sgn-base-03 rotate: F subst: [bits, boolvar] → [bits0, boolvar1]

Additional information

Used terms: bits0, boolvar1