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Proof step 1 from proof replay-lpt-at (conclusion)

m < n, n ≤ # lpt ⊦ replay-lpt(lpt, nx, n)[m] = (lpt[m].flags = LP_FREE ∨ m ∈ nx ⊃ lpt[m] ;(lpt[m].flags = LP_INDEX_NODES ∧ lpt[m].ref-size ≠ 0 ⊃ lpt[m].size:= LEB_SIZE ;(lpt[m].ref-size = 0 ⊃ mklp(0, 0, LP_FREE, lpt[m].gcheapidx);lpt[m]) ) )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < n, m < # lpt, n ≤ # lpt ⊦ replay-lpt(lpt, nx, n)[m] = (lpt[m].flags = LP_FREE ∨ m ∈ nx ⊃ lpt[m] ;(lpt[m].flags = LP_INDEX_NODES ∧ lpt[m].ref-size ≠ 0 ⊃ lpt[m].size:= LEB_SIZE ;(lpt[m].ref-size = 0 ⊃ mklp(0, 0, LP_FREE, lpt[m].gcheapidx);lpt[m]) ) )

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic