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Proof step 7 from proof sgn-in-01 (previous step)

∀ boolvar. boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0 ⊦ boolvar0 ∈ sgn(boolvar ' + bits0) ↔ (boolvar0 ↔ boolvar) ∨ boolvar0 ∈ bits0, bits0 = [], bits0.first ↔ boolvar

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
bits0 = boolvar1 + bits, ∀ boolvar. boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0 ⊦ boolvar0 ∈ sgn(boolvar ' + bits0) ↔ (boolvar0 ↔ boolvar) ∨ boolvar0 ∈ bits0, bits0 = [], boolvar1 ↔ boolvar

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar1, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits0, boolvar1, bits