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Proof step 18 from proof key-set-write-nodups (previous step)

n < m0, m0 < # kar, n < # kar, n + 1 < # kar, ∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m].idx ≠ kar[m1].idx, ∀ n, kar, ki. n < m0 ∧ n < # kar ∧ m0 ≤ # kar ∧ (∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2].idx ≠ kar[m1].idx) → keyset(kar[n, ki], m0) = (keyset(kar, m0) -- kar[n]) ++ ki ⊦ keyset(kar[n, ki], m0) ++ kar[m0] = ((keyset(kar, m0) ++ kar[m0]) -- kar[n]) ++ ki, m0 = n

User applied rule all left

And obtained the following new goals

Proof step 19:
n + 1 < # kar, n < # kar, m0 < # kar, n < m0, n < m0 ∧ n < # kar ∧ m0 ≤ # kar ∧ (∀ m2, m1. m2 < # kar ∧ m2 < m0 ∧ m1 < # kar ∧ m1 < m0 ∧ m2 ≠ m1 → kar[m2].idx ≠ kar[m1].idx) → keyset(kar[n, ki], m0) = (keyset(kar, m0) -- kar[n]) ++ ki, ∀ m, m1. m < # kar ∧ ¬ m0 < m ∧ m1 < # kar ∧ ¬ m0 < m1 ∧ m ≠ m1 → kar[m].idx ≠ kar[m1].idx ⊦ keyset(kar[n, ki], m0) ++ kar[m0] = ((keyset(kar, m0) ++ kar[m0]) -- kar[n]) ++ ki, m0 = n

Rule argument

pos: left 6 terms: n, kar, ki abort: F discard: T computed: F

Additional information

Used terms: n, kar, ki