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Proof step 4 from proof sgn-append-01 (previous step)

false ∈ bits0 ∧ true ∈ bits0 → (∀ bits1. sgn(bits0 + bits1) = sgn(bits0) + bits1) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn(boolvar ' + bits0) + bits1, ¬ false ∈ bits0 ∧ boolvar, ¬ true ∈ bits0 ∧ ¬ 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)), false ∈ bits0 ∧ true ∈ bits0 → (∀ bits1. sgn(bits0 + bits1) = sgn(bits0) + bits1) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn(boolvar ' + bits0) + bits1, ¬ false ∈ bits0 ∧ boolvar, ¬ true ∈ bits0 ∧ ¬ boolvar
Proof step 11:
bits0 ≠ [] ∧ (bits0.first ↔ boolvar), false ∈ bits0 ∧ true ∈ bits0 → (∀ bits1. sgn(bits0 + bits1) = sgn(bits0) + bits1) ⊦ sgn((boolvar ' + bits0) + bits1) = sgn(bits0) + bits1, ¬ false ∈ bits0 ∧ boolvar, ¬ true ∈ bits0 ∧ ¬ boolvar

Rule argument

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

Additional information

Used terms: bits0, boolvar