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Proof step 26 from proof prefix-at (previous step)

# x = # lax2 + # lax1 + 1, # x0 = # lax2 + (# lax1 + # lax) + 1, st1[m] = mkblock(x0, (lax2 + @(m0, m1) ' + lax1) + lax, m3), st0[m] = mkblock(x, lax2 + @(m0, m1) ' + lax1, m2), m ∈ st1, m ∈ st0, x ⊑ x0, ∀ n. n ∈ st0 → n ∈ st1 ⊦ x[# lax2] = x0[# lax2], @(m0, m1) ∈ lax2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 28:
# x0 = # lax2 + (# lax1 + # lax) + 1, # x = # lax2 + # lax1 + 1, st0[m] = mkblock(x, lax2 + @(m0, m1) ' + lax1, m2), st1[m] = mkblock(x0, (lax2 + @(m0, m1) ' + lax1) + lax, m3), m ∈ st0, m ∈ st1, ∀ n. n ∈ st0 → n ∈ st1, ∃ z. x + z = x0 ⊦ x[# lax2] = x0[# lax2], @(m0, m1) ∈ lax2

Rule argument

spec: list-last name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [x0, x] flags: <all paths>

Additional information

Used terms: x0, x