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Proof step 11 from proof ri-cons-copy-new (previous step)

# kx1 + # kx0 + 1 = # ax, new(ax, fs), key ∈ ri, kx0.toSet ⊆ dom ri, kx1.toSet ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ dups(ax), dups(kx1 + kx0), dups(kx1), dups(kx0), key ∈ kx1, key ∈ kx0, (ax[# kx1] ∈ fs ∨ ax[# kx1] ∈ ax) ∧ key-cons(key, fs[ax, fs[ri[kx1]] + fs[ri[key]] ' + fs[ri[kx0]]][ax[# kx1]])

User applied rule apply lemma

And obtained the following new goals

Proof step 12:
n ≤ # ax ⊦ ∃ ax0, ax1. ax = ax0 + ax1 ∧ # ax0 = n
Proof step 13:
# kx1 + # kx0 + 1 = # ax, new(ax, fs), key ∈ ri, kx1.toSet ⊆ dom ri, kx0.toSet ⊆ dom ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∃ ax0, ax1. ax = ax0 + ax1 ∧ # ax0 = # kx1 ⊦ dups(kx0), dups(kx1), dups(kx1 + kx0), dups(ax), key ∈ kx0, key ∈ kx1, (ax[# kx1] ∈ fs ∨ ax[# kx1] ∈ ax) ∧ key-cons(key, fs[ax, fs[ri[kx1]] + fs[ri[key]] ' + fs[ri[kx0]]][ax[# kx1]])

Rule argument

spec: list, inst: address-list-set name: split-first seq: n ≤ # ax ⊦ ∃ ax0, ax1. ax = ax0 + ax1 ∧ # ax0 = n subst: [n, ax] → [# kx1, ax]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ # kx = 0 ↔ kx = []; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: # kx1, ax