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Proof step 102 from proof prepost-remove (previous step)

# kar0 = # kar, keyset(kar, m2) ++ key-index(key0, m0) = keyset(kar0, m2), ar1[m0] = elemindex(a, m1), kar[m1] = key-index(key0, m0), kar[m2] = key-index(key0, m0), kar[2 * m1 + 1] = key-index(key2, m4), kar[2 * m1 + 2] = key-index(key3, m5), kar[(m1 - 1) / 2] = key-index(key1, m3), 2exp-size(# kar), m1 + 1 < # kar, m1 < # kar, m2 < # kar, m0 < # ar1, m1 < m2, heap-order(kar, m2 + 1), heap-order(kar0, m2), heap-order(kar, m2), id-idx(ar1, ar0), no-dups-idx(kar0, m2), valid-idx(kar, ar1, m2 + 1), valid-idx(kar0, ar0, m2), 2 * m1 + 1 < m2 ∧ key2 < key0 ∨ 2 * m1 + 2 < m2 ∧ key3 < key0 → heap-order(kar[m1, (2 * m1 + 2 < m2 ∧ key3 < key2 ⊃ key-index(key3, m5);key-index(key2, m4))], m2), elemindex(a, m1) = elemindex(a, m3) ∧ valid-idx(kar, ar1, m1 + 1) → kar[m3] = key-index(key0, m0), m1 ≠ 0 → (m1 - 1) / 2 < # kar ∧ 1 ≤ m1, ∀ m, m6. m < # kar ∧ ¬ m2 < m ∧ m6 < # kar ∧ ¬ m2 < m6 ∧ m ≠ m6 → kar[m].idx ≠ kar[m6].idx ⊦ #(keyset(kar, m2) ++ key-index(key0, m0)) = 0, m2 = m1, m1 ≠ 0 ∧ key0 < key1

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)% ⊦ ¬ n < n; (s) Used terms: m1, m2