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Proof step 5 from proof cleanup-lpt-n-size (previous step)

lpt[n0] = mklp(m, m0, lpf, m1), n0 < # lpt, ∀ lpt. n0 ≤ # lpt → # cleanup-lpt(lpt, n0) = # lpt ⊦ # (m = 0 ∧ lpf ≠ LP_FREE ⊃ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m1)];cleanup-lpt(lpt, n0)) = # lpt

User applied rule case distinction

And obtained the following new goals

Proof step 6:
lpt[n0] = mklp(m, m0, lpf, m1), m = 0, n0 < # lpt, ∀ lpt. n0 ≤ # lpt → # cleanup-lpt(lpt, n0) = # lpt ⊦ # cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m1)] = # lpt, lpf = LP_FREE
Proof step 8:
lpt[n0] = mklp(m, m0, lpf, m1), n0 < # lpt, ∀ lpt. n0 ≤ # lpt → # cleanup-lpt(lpt, n0) = # lpt, m ≠ 0 ∨ lpf = LP_FREE ⊦ # cleanup-lpt(lpt, n0) = # lpt

Rule argument

pos: right 1

Proof step 5 from proof cleanup-lpt-n-size (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information