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Proof step 21 from proof bin-heap-inv-cleanup-lpt (previous step)

kar[m0] = key-index(natkey, m4), lpt[m4] = mklp(m1, m2, lpf, m3), 2exp-size(# kar), m0 < # kar, m0 < m, heap-order(kar, m), no-dups-idx(kar, m), m ≤ # kar, m0 < m → kar[m0].idx < # lpt ∧ lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[kar[m0].idx].gcheapidx = m0 ⊦ m4 < # lpt ∧ m3 = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
lpt[m4] = mklp(m1, m2, lpf, m3), kar[lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[m4].gcheapidx] = key-index(natkey, m4), 2exp-size(# kar), m4 < # lpt, lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[m4].gcheapidx < m, lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[m4].gcheapidx < # kar, heap-order(kar, m), no-dups-idx(kar, m), m ≤ # kar ⊦ m3 = lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[m4].gcheapidx

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ key-index(natkey, m).idx = m; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic