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Proof step 6 from proof orderedls-ls (
previous step
)
b < a0, ordered<(a0 ' + x0), a < a0, b < a, b ∈ x0, ∀ a1, b0. ¬ ordered<(a1 ' + x0) ∨ ¬ b0 < a1 ∨ ¬ b0 ∈ x0 ⊦ b = a0
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 6 terms: a0, b
Additional information
Used terms: a0, b