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Proof step 55 from proof prepost-remove (previous step)

# kar0 = # kar, keyset(kar, m1) ++ key-index(key0, m0) = keyset(kar0, m1), kar[m1] = key-index(key0, m0), ar1[m0] = elemindex(a, m2 + 1), kar[m2 + 1] = key-index(key0, m0), kar[m2 / 2] = key-index(key1, m3), 2exp-size(# kar), m2 / 2 < # kar, key0 < key1, m2 + 1 < # kar, m1 < # kar, m0 < # ar1, m2 < m1, m2 < # kar, 2 + m2 < # kar, m2 + 1 < m1, heap-order(kar0, m1), heap-order(kar, m1 + 1), heap-order(kar[m2 + 1, key-index(key1, m3)], m1), id-idx(ar1, ar0), no-dups-idx(kar0, m1), valid-idx(kar0, ar0, m1), valid-idx(kar, ar1, m1 + 1), elemindex(a, m2 + 1) = elemindex(a, m0 + 1) → kar[m0 + 1] = key-index(key0, m0), ∀ m, m4. m < # kar ∧ ¬ m1 < m ∧ m4 < # kar ∧ ¬ m1 < m4 ∧ m ≠ m4 → kar[m].idx ≠ kar[m4].idx ⊦ # kar = 1, #(keyset(kar, m1) ++ key-index(key0, m0)) = 0, m1 = m2 + 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 25 terms: m1, m2 + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m1, m2 + 1