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Proof step 36 from proof bin-heap-inv-cleanup-lpt (previous step)

(m0 = 0 ∧ lpf0 ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m6);mklp(m0, m5, lpf0, m6)) = mklp(m1, m2, lpf, m3), lpt[m4] = mklp(m0, m5, lpf0, m6), kar[m6] = key-index(natkey, m4), 2exp-size(# kar), m6 < # kar, m4 < # lpt, m6 < m, heap-order(kar, m), no-dups-idx(kar, m), m ≤ # kar ⊦ m3 = m6

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ mklp(m, m0, lpf, m1) = mklp(m2, m3, lpf0, m4) ↔ m = m2 ∧ m0 = m3 ∧ lpf = lpf0 ∧ m1 = m4; (ws)