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Proof step 7 from proof lpt-leb-size-id-idx (previous step)

# lpt1 = # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags, ∀ n. n < # lpt0 → lpt1[n].size ≤ LEB_SIZE ⊦ ∀ n. n < # lpt0 → lpt0[n].size ≤ LEB_SIZE

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# lpt1 = # lpt0, LEB_SIZE < lpt0[n].size, n < # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags, ∀ n. n < # lpt0 → lpt1[n].size ≤ LEB_SIZE ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)