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Proof step 5 from proof restr-add-01 (previous step)

∀ as, m0, m1, m2. (as ++ @(m0, m1, m2)) | m3 = (as | m3) ∪ (m0 = m3 ⊃ ∅ ++ @(m3, m1, m2);∅) ⊦ (as ++ @(m0, m1, m2)) | m3 +1 = (as | m3 +1) ∪ (m0 = m3 +1 ⊃ ∅ ++ @(m3 +1, m1, m2);∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ as, m0, m1, m2. (as ++ @(m0, m1, m2)) | m3 = (as | m3) ∪ (m0 = m3 ⊃ ∅ ++ @(m3, m1, m2);∅) ⊦ (as ++ @(m0, m1, m2)) | (m3 + 1) = (as | (m3 + 1)) ∪ (m0 = m3 + 1 ⊃ ∅ ++ @(m3 + 1, m1, m2);∅)

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)