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Proof step 4 from proof freelist-cons-add (previous step)

lpt[n] = mklp(n0, n1, lpf, n2), lpt[n, mklp(m0, m1, LP_FREE, m3)][n3] = mklp(m, m2, lpf0, m4), n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, dups(freelist), ¬ n ∈ freelist ∧ (n3 < # lpt ∧ lpf0 = LP_FREE ↔ n3 ∈ freelist ∨ n3 = n)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
lpt[n, mklp(m0, m1, LP_FREE, m3)][n3] = mklp(m, m2, lpf0, m4), lpt[n] = mklp(n0, n1, lpf, n2), n3 = n, n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, dups(freelist), ¬ n ∈ freelist ∧ (n3 < # lpt ∧ lpf0 = LP_FREE ↔ n3 ∈ freelist ∨ n3 = n)
Proof step 7:
lpt[n, mklp(m0, m1, LP_FREE, m3)][n3] = mklp(m, m2, lpf0, m4), lpt[n] = mklp(n0, n1, lpf, n2), n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, n3 = n, dups(freelist), ¬ n ∈ freelist ∧ (n3 < # lpt ∧ lpf0 = LP_FREE ↔ n3 ∈ freelist ∨ n3 = n)

Rule argument

fma: n3 = n

Additional information

Used formulas: n3 = n