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Proof step 4 from proof to-adrs-eq-app (previous step)

@(n, ladr.off, ladr.size) ' + lax1.to-adr(n) = lax.to-adr(n) + @(n, m0, m1) ', ∀ n, lax, m0, m1. lax1.to-adr(n) = lax.to-adr(n) + @(n, m0, m1) ' ∧ @(m0, m1) ∈ lax1 → lax1 = lax + @(m0, m1) ', @(m0, m1) = ladr ∨ @(m0, m1) ∈ lax1, @(n, ladr.off, ladr.size) ∈ lax.to-adr(n) ∨ @(n, ladr.off, ladr.size) = @(n, m0, m1), @(n, m0, m1) = @(n, ladr.off, ladr.size) ∨ @(n, m0, m1) ∈ lax1.to-adr(n) ⊦ ladr ' + lax1 = lax + @(m0, m1) '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
@(n, m, m2) ' + lax1.to-adr(n) = lax.to-adr(n) + @(n, m0, m1) ', ladr = @(m, m2), ∀ n, lax, m0, m1. lax1.to-adr(n) = lax.to-adr(n) + @(n, m0, m1) ' ∧ @(m0, m1) ∈ lax1 → lax1 = lax + @(m0, m1) ', @(n, m0, m1) = @(n, m, m2) ∨ @(n, m0, m1) ∈ lax1.to-adr(n), @(n, m, m2) ∈ lax.to-adr(n) ∨ @(n, m, m2) = @(n, m0, m1), @(m0, m1) = ladr ∨ @(m0, m1) ∈ lax1 ⊦ ladr ' + lax1 = lax + @(m0, m1) '

Rule argument

spec: laddress name: elim seq: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0) subst: [ladr, m, m0] → [ladr, m, m2]

Additional information

Used simplifier rules: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0); (elim) Used terms: ladr, m, m2