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Proof step 15 from proof range-del-in (previous step)

key0 ∈ ri, key ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ key0 = key, ri[key0] ≠ ri[key] ∧ (∃ key. key ∈ ri ∧ ri[key] = ri[key0])

User applied rule all left

And obtained the following new goals

Proof step 16:
key ∈ ri, key0 ∈ ri, key0 ∈ ri ∧ key ∈ ri ∧ ri[key0] = ri[key] → key0 = key, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ key0 = key, ri[key0] ≠ ri[key] ∧ (∃ key. key ∈ ri ∧ ri[key] = ri[key0])

Rule argument

pos: left 3 terms: key0, key abort: F discard: F computed: T

Additional information

Used terms: key0, key