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Proof step 5 from proof ri-inj-index-update (previous step)

# ax = # kx, ri[kx, ax][key] = ri[kx, ax][key0], ax.toSet ∩ range ri = ∅, ri-inj(ri), key0 ∈ ri[kx, ax], key ∈ ri[kx, ax], ∀ kx0, ax, ri, key, key0. # kx0 < # kx → # ax = # kx0 ∧ ri[kx0, ax][key] = ri[kx0, ax][key0] ∧ ax.toSet ∩ range ri = ∅ ∧ ri-inj(ri) ∧ key ∈ ri[kx0, ax] ∧ key0 ∈ ri[kx0, ax] → key = key0 ∨ dups(ax) ∨ key ∈ ri ∧ key0 ∈ ri ∧ ri[key] = ri[key0] ⊦ key = key0, dups(ax), key ∈ ri ∧ key0 ∈ ri ∧ ri[key] = ri[key0]

User applied rule cut formula

And obtained the following new goals

Proof step 6:
# ax = # kx, ri[kx, ax][key] = ri[kx, ax][key0], ax.toSet ∩ range ri = ∅, ri-inj(ri), key ∈ ri[kx, ax], key0 ∈ ri[kx, ax], ∀ kx0, ax, ri, key, key0. # kx0 < # kx → # ax = # kx0 ∧ ri[kx0, ax][key] = ri[kx0, ax][key0] ∧ ax.toSet ∩ range ri = ∅ ∧ ri-inj(ri) ∧ key ∈ ri[kx0, ax] ∧ key0 ∈ ri[kx0, ax] → key = key0 ∨ dups(ax) ∨ key ∈ ri ∧ key0 ∈ ri ∧ ri[key] = ri[key0] ⊦ key = key0, kx = [], dups(ax), key ∈ ri ∧ key0 ∈ ri ∧ ri[key] = ri[key0]

Rule argument

fma: kx = []

Additional information

Used simplifier rules: ⊦ # [] = 0; (s)% ⊦ # ax = 0 ↔ ax = []; (s)% ⊦ ri[[], []] = ri; (s)% ⊦ [].toSet = ∅; (s)% ⊦ as ∩ ∅ = ∅; (s)%int-a: ⊦ (as ∩ as0) ∩ as1 = as ∩ as0 ∩ as1; from specification set-union[node-store]%int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[node-store]% ⊦ ¬ dups([]); (s) Used formulas: kx = []