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Proof step 21 from proof ladrlist-to-adr-split (previous step)

¬ # lax3 + # lax < # lax3 → ( (lax3.to-adr(m) + @(m, m1, m2) ' + lax.to-adr(m))[# lax3] = @(m, m1, m2) ↔ (∃ lax2, lax1. lax3 + @(m1, m2) ' + lax = lax2 + @(m1, m2) ' + lax1 ∧ m = m ∧ # lax2 = # lax3)), ∀ n, m0, m3, m4, m5. ¬ # lax3 + # lax < n → ( (lax3.to-adr(m0) + @(m0, m1, m2) ' + lax.to-adr(m0))[n] = @(m3, m4, m5) ↔ (∃ lax2, lax1. lax3 + @(m1, m2) ' + lax = lax2 + @(m4, m5) ' + lax1 ∧ m3 = m0 ∧ # lax2 = n)) ⊦ (lax3.to-adr(m) + @(m, m1, m2) ' + lax.to-adr(m))[# lax3] = @(m, m1, m2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
∀ n, m0, m3, m4, m5. ¬ # lax3 + # lax < n → ( (lax3.to-adr(m0) + @(m0, m1, m2) ' + lax.to-adr(m0))[n] = @(m3, m4, m5) ↔ (∃ lax2, lax1. lax3 + @(m1, m2) ' + lax = lax2 + @(m4, m5) ' + lax1 ∧ m3 = m0 ∧ # lax2 = n)) ⊦ (lax3.to-adr(m) + @(m, m1, m2) ' + lax.to-adr(m))[# lax3] = @(m, m1, m2), ∃ lax0, lax1. lax3 + @(m1, m2) ' + lax = lax0 + @(m1, m2) ' + lax1 ∧ # lax0 = # lax3

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)