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Proof step 6 from proof to-adrs-in (previous step)

@(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ @(m0, m1) ∈ lax0 ⊦ @(m, m0, m1) = @(n, m2, m3) ∨ @(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ (@(m0, m1) = @(m2, m3) ∨ @(m0, m1) ∈ lax0)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
@(m, m0, m1) ∈ lax0.to-adr(n) ↔ m = n ∧ @(m0, m1) ∈ lax0, @(m, m0, m1) = @(n, m2, m3) ∨ @(m, m0, m1) ∈ lax0.to-adr(n) ⊦ m = n ∧ (@(m0, m1) = @(m2, m3) ∨ @(m0, m1) ∈ lax0)

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ @(m, m0) = @(m1, m2) ↔ m = m1 ∧ m0 = m2; (ws)