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Proof step 9 from proof del1-01-02 (
previous step
)
a0 < a, ordered≤(a0 ' + x0), ∀ a1, b0. ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ∨ a1 = b0 ⊦ b = a0, a = b, ins≤(a, x0 -1l b) = ins≤(a, x0) -1l b
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: a, b
Additional information
Used terms: a, b