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Proof step 3 from proof ri-inj-variant1-01 (previous step)

∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ (∀ key0, key1. key0 ∈ ri0 ∧ key1 ∈ ri0 ∧ ri0[key0] = ri0[key1] → key0 = key1) ↔ (∀ key, key0. ¬ key ∈ ri ∨ ¬ key0 ∈ ri ∨ ri[key] ≠ ri[key0] → key ∈ ri0 ∧ key0 ∈ ri0 ∧ ri0[key] = ri0[key0] → key = key0)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1, ∀ key0, key1. key0 ∈ ri0 ∧ key1 ∈ ri0 ∧ ri0[key0] = ri0[key1] → key0 = key1 ⊦ ∀ key, key0. ¬ key ∈ ri ∨ ¬ key0 ∈ ri ∨ ri[key] ≠ ri[key0] → key ∈ ri0 ∧ key0 ∈ ri0 ∧ ri0[key] = ri0[key0] → key = key0
Proof step 6:
∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1, ∀ key, key0. ¬ key ∈ ri ∨ ¬ key0 ∈ ri ∨ ri[key] ≠ ri[key0] → key ∈ ri0 ∧ key0 ∈ ri0 ∧ ri0[key] = ri0[key0] → key = key0 ⊦ ∀ key0, key1. key0 ∈ ri0 ∧ key1 ∈ ri0 ∧ ri0[key0] = ri0[key1] → key0 = key1

Rule argument

pos: right 1

Proof step 3 from proof ri-inj-variant1-01 (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information