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Proof step 10 from proof sgn-in-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ boolvar. boolvar ∈ sgn(bits0) ↔ boolvar ∈ bits0 ⊦ boolvar0 ∈ sgn(bits0) ↔ (boolvar0 ↔ bits0.first) ∨ boolvar0 ∈ bits0, bits0 = []

Proof step 10 from proof sgn-in-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information