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Proof step 13 from proof sgn-in-01 (
previous step
)
∀ boolvar. boolvar ∈ sgn(boolvar1 ' + bits) ↔ boolvar ∈ bits ∨ (boolvar ↔ boolvar1) ⊦ boolvar0 ∈ sgn(boolvar1 ' + bits) ↔ boolvar0 ∈ bits ∨ (boolvar0 ↔ boolvar1)
User applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: boolvar0 abort: F discard: T computed: T
Additional information
Used terms: boolvar0