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

@(n, m, m0) ' + lax.to-adr(n) = lax1.to-adr(n), @(m, m0) ∈ lax1, ∀ n, lax1. lax.to-adr(n) = lax1.to-adr(n) → lax = lax1 ⊦ @(m, m0) ' + lax = lax1

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
@(n, m, m0) ' + lax.to-adr(n) = lax1.to-adr(n), lax1 = ladr + lax0, @(m, m0) ∈ lax1, ∀ n, lax1. lax.to-adr(n) = lax1.to-adr(n) → lax = lax1 ⊦ @(m, m0) ' + lax = lax1

Rule argument

var: lax1

Additional information

Used simplifier rules: adrs-empty: ⊦ ([]).to-adr(n) = [];%laddress-list freely generated by [], +;;% ⊦ ax + ax0 = [] ↔ ax = [] ∧ ax0 = []; (as)% ⊦ adr ' ≠ []; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]