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Proof step 7 from proof diff-consnotin2 (
previous step
)
a0 ∈ x0, ∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ a = a0, x1 \ (a ' + x0) = x1 \ x0, a ∈ x1
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: a, x0
Additional information
Used terms: a, x0