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Proof step 4 from proof size-rec-rem (previous step)

⊦ (as ++ @(m, m0, m1)).size = (as.size - (@(m, m0, m1) ∈ as ⊃ m1;0)) + m1

User applied rule case distinction

And obtained the following new goals

Proof step 5:
@(m, m0, m1) ∈ as ⊦ (as ++ @(m, m0, m1)).size = (as.size - m1) + m1

Rule argument

pos: right 1

Additional information

Used simplifier rules: size-rec: ⊦ ¬ adr ∈ as → (as ++ adr).size = as.size + adr.size;% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ m - 0 = m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic