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Proof step 9 from proof index-update-range (previous step)

# kx0 + 1 = # ax, ax.toSet ∩ range ri = ∅, ri-inj(ri), key ∈ ri, kx0.toSet ⊆ dom ri, ∀ kx1, ax, ri. # kx1 < #(kx0 + key ') → # kx1 = # ax ∧ ax.toSet ∩ range ri = ∅ ∧ ri-inj(ri) ∧ kx1.toSet ⊆ dom ri → range ri[kx1, ax] = (range ri \ ri[kx1].toSet) ∪ ax.toSet ∨ dups(kx1) ∨ dups(ax) ⊦ range ri[kx0 + key ', ax] = ((range ri \ ri[kx0].toSet) -- ri[key]) ∪ ax.toSet, dups(ax), dups(kx0), key ∈ kx0

User applied rule cut formula

And obtained the following new goals

Proof step 10:
# kx0 + 1 = # ax, ax.toSet ∩ range ri = ∅, ri-inj(ri), key ∈ ri, kx0.toSet ⊆ dom ri, ∀ kx1, ax, ri. # kx1 < #(kx0 + key ') → # kx1 = # ax ∧ ax.toSet ∩ range ri = ∅ ∧ ri-inj(ri) ∧ kx1.toSet ⊆ dom ri → range ri[kx1, ax] = (range ri \ ri[kx1].toSet) ∪ ax.toSet ∨ dups(kx1) ∨ dups(ax) ⊦ ax = [], range ri[kx0 + key ', ax] = ((range ri \ ri[kx0].toSet) -- ri[key]) ∪ ax.toSet, dups(kx0), dups(ax), key ∈ kx0

Rule argument

fma: ax = []

Additional information

Used simplifier rules: ⊦ [].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]% ⊦ # [] = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ # kx = 0 ↔ kx = []; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: ax = []