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Proof step 5 from proof to-adrs-eq-one (previous step)

@(m, m0, m1) ' = lax.to-adr(n), lax = ladr + lax0 ⊦ lax = @(m0, m1) ' ∧ m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ @(ladr.off, ladr.size) = ladr

Rule argument

Additional information

Used simplifier rules: adrs-rec: ⊦ (ladr ' + lax).to-adr(n) = @(n, ladr.off, ladr.size) + lax.to-adr(n);%to-adrs-eq-empty: ⊦ lax.to-adr(n) = [] ↔ lax = [];% ⊦ ladr ' = ladr0 ' + lax ↔ ladr = ladr0 ∧ lax = []; (ws)% ⊦ @(m, m0, m1) = @(m2, m3, m4) ↔ m = m2 ∧ m0 = m3 ∧ m1 = m4; (ws)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ adr ' = adr0 ' + ax ↔ adr = adr0 ∧ ax = []; (ws)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ adr + ax = adr ' + ax; (s)% ⊦ ladr + lax = ladr ' + lax; (s)