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Proof step 15 from proof in-ri-update-in-ri (previous step)

# ax1 = # ax0, adr ∈ ri, ax1.toSet ⊆ range ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), key = fs[adr].key ∨ key ∈ keys(fs, ax1) ⊦ key ∈ ri, key ∈ ri[keys(fs, ax1), ax1]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ # kx = # ax → (key ∈ ri[kx, ax] ↔ key ∈ ri ∨ key ∈ kx)
Proof step 17:
# ax1 = # ax0, # keys(fs, ax1) = # ax1, adr ∈ ri, ax1.toSet ⊆ range ri, ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), key = fs[adr].key ∨ key ∈ keys(fs, ax1) ⊦ key ∈ keys(fs, ax1), key ∈ ri

Rule argument

spec: aubifs-base name: index-update-in seq: ⊦ # kx = # ax → (key ∈ ri[kx, ax] ↔ key ∈ ri ∨ key ∈ kx) subst: [ax, kx, ri, key] → [ax1, keys(fs, ax1), ri, key] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # keys(fs, ax) = # ax; (s) Used terms: ax1, keys(fs, ax1), ri, key