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Proof step 68 from proof prepost-extract-min (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 18 terms: m0, 0 abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m < 0; (s) Used terms: m0, 0