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Proof step 14 from proof Del-at (
previous step
)
∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0 -- b)[a] = st0[a], a = a0, a = b, b = a0
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: b, a
Additional information
Used terms: b, a