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Proof step 5 from proof lpt-free-unref-write (previous step)

∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ (∀ n0. n0 < # lpt ∧ lpt[n, mklp(m, m0, lpf, m1)][n0].flags = LP_FREE → lpt[n, mklp(m, m0, lpf, m1)][n0].ref-size = 0) ↔ n < # lpt ∧ lpf = LP_FREE → m = 0

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 6:
n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ (∀ n0. n0 < # lpt ∧ lpt[n, mklp(m, m0, lpf, m1)][n0].flags = LP_FREE → lpt[n, mklp(m, m0, lpf, m1)][n0].ref-size = 0) ↔ n < # lpt ∧ lpf = LP_FREE → m = 0

Rule argument

pos: left 1 terms: n

Additional information

Used terms: n