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Proof step 17 from proof lpt-ram-flash-id-idx (previous step)

# lpt = # lpt0, # 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 ∧ lpt[n].flags = LP_GROUP_NODES ∧ lpt0[n].flags = LP_GROUP_NODES ∧ ¬ n ∈ nx → lpt0[n].size = lpt[n].size ⊦ ∀ n. n < # lpt0 ∧ lpt[n].flags = LP_GROUP_NODES ∧ lpt1[n].flags = LP_GROUP_NODES ∧ ¬ n ∈ nx → lpt1[n].size = lpt[n].size

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# lpt = # lpt0, # lpt1 = # lpt0, lpt1[n].flags = LP_GROUP_NODES, lpt[n].flags = LP_GROUP_NODES, 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 ∧ lpt[n].flags = LP_GROUP_NODES ∧ lpt0[n].flags = LP_GROUP_NODES ∧ ¬ n ∈ nx → lpt0[n].size = lpt[n].size ⊦ lpt1[n].size = lpt[n].size, n ∈ nx

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