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Proof step 36 from proof ri-inj-index-update (previous step)

# ax0 = # kx0, ri[kx0, ax0][key] = ri[kx0, ax0][key0], range ri ∩ ax0.toSet = ∅, ri-inj(ri), key ∈ ri[kx0, ax0], key0 ∈ ri[kx0, ax0], ∀ kx1, ax, ri, key, key0. # kx1 < #(kx0 + key1 ') → # ax = # kx1 ∧ ri[kx1, ax][key] = ri[kx1, ax][key0] ∧ ax.toSet ∩ range ri = ∅ ∧ ri-inj(ri) ∧ key ∈ ri[kx1, ax] ∧ key0 ∈ ri[kx1, ax] → key = key0 ∨ dups(ax) ∨ key ∈ ri ∧ key0 ∈ ri ∧ ri[key] = ri[key0] ⊦ key = key1, key = key0, key0 = key1, dups(ax0), adr ∈ ri, adr ∈ ax0, key ∈ ri ∧ key0 ∈ ri ∧ ri[key] = ri[key0]

User applied rule apply induction

And closed the goal!

Rule argument

subst: [kx1, ax, ri, key, key0] → [kx0, ax0, ri, key, key0]

Additional information

Used simplifier rules: int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[node-store]% ⊦ #(kx + kx0) = # kx + # kx0; (s)% ⊦ # key ' = 0 +1; (s)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: kx0, ax0, ri, key, key0