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Proof step 3 from proof restr-add-01 (previous step)

⊦ (as ++ @(m0, m1, m2)) | m = (as | m) ∪ (m0 = m ⊃ ∅ ++ @(m, m1, m2);∅)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ (as ++ @(m0, m1, m2)) | 0 = (as | 0) ∪ (m0 = 0 ⊃ ∅ ++ @(0, m1, m2);∅)
Proof step 5:
∀ 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

term: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;