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Proof step 3 from proof mk-inj (previous step)

∀ a, b, n. mklist(a, m0) = mklist(b, n) → m0 = 0 ∨ a = b ∧ m0 = n, mklist(a, m0 +1) = mklist(b, n) ⊦ m0 +1 = 0, a = b ∧ m0 +1 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b ' + mklist(b, m0) = mklist(b, n), ∀ a, b, n. mklist(a, m0) = mklist(b, n) → m0 = 0 ∨ a = b ∧ m0 = n ⊦ n = 0, m0 + 1 = n

Rule argument

Additional information

Used simplifier rules: mk-plus: ⊦ mklist(a, m + n) = mklist(a, m) + mklist(a, n);%mklist-one: ⊦ mklist(a, 1) = a ';%mk-last: ⊦ mklist(a, n) + a ' = a ' + mklist(a, n);%mk-elem: ⊦ a ∈ mklist(b, n) → a = b;%mk-in: ⊦ a ∈ mklist(a, m) ↔ m ≠ 0;%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2