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Proof step 4 from proof sgn-notsame (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
bits ≠ [], bits.first ↔ boolvar ⊦ sgn(boolvar ' + bits) = sgn(bits)
Proof step 6:
¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), sgn(boolvar ' + bits0) = boolvar0 ' + boolvar0 ', ∀ boolvar1. sgn(bits0) ≠ boolvar1 ' + boolvar1 ' ∨ ¬ boolvar1 ∈ sgn(bits0) ⊦ ¬ boolvar0 ∈ bits0 ∧ ¬(boolvar0 ↔ boolvar)
Proof step 10:
bits0 ≠ [] ∧ (bits0.first ↔ boolvar), sgn(bits0) = boolvar0 ' + boolvar0 ', ∀ boolvar1. sgn(bits0) ≠ boolvar1 ' + boolvar1 ' ∨ ¬ boolvar1 ∈ sgn(bits0) ⊦ ¬ boolvar0 ∈ bits0 ∧ ¬(boolvar0 ↔ boolvar)

Rule argument

name: sgn-rec-01 rotate: F subst: [bits, boolvar] → [bits0, boolvar]

Additional information

Used terms: bits0, boolvar