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Proof step 19 from proof union-write (previous step)

n < m0, m0 < # asar, n < # asar, n + 1 < # asar, disj(asar, 0, m0 + 1), ∀ n, asar, as. n < m0 ∧ n < # asar ∧ disj(asar, 0, m0) ∧ m0 ≤ # asar → union(asar[n, as], 0, m0) = (union(asar, 0, m0) \ asar[n]) ∪ as ⊦ m0 = n, asar[m0] ∪ union(asar[n, as], 0, m0) = (asar[m0] ∪ (union(asar, 0, m0) \ asar[n])) ∪ as

User applied rule all left

And obtained the following new goals

Proof step 20:
n + 1 < # asar, n < # asar, m0 < # asar, n < m0, disj(asar, 0, m0 + 1), n < m0 ∧ n < # asar ∧ disj(asar, 0, m0) ∧ m0 ≤ # asar → union(asar[n, as], 0, m0) = (union(asar, 0, m0) \ asar[n]) ∪ as ⊦ m0 = n, asar[m0] ∪ union(asar[n, as], 0, m0) = (asar[m0] ∪ (union(asar, 0, m0) \ asar[n])) ∪ as

Rule argument

pos: left 6 terms: n, asar, as abort: F discard: T computed: F

Additional information

Used terms: n, asar, as