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Proof step 4 from proof from-rest (previous step)

a ∈ a0 + x0, ∀ a. a ∈ x0 → frome(x0, a) = restn(pos(a, x0), x0) ⊦ frome(a0 + x0, a) = restn(pos(a, a0 + x0), a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x0, ∀ a. a ∈ x0 → frome(x0, a) = restn(pos(a, x0), x0) ⊦ a = a0, frome(x0, a) = restn(pos(a, x0), x0)

Rule argument

Additional information

Used simplifier rules: fromE-yes: ⊦ frome(a ' + x, a) = a ' + x;%pos-y: ⊦ pos(a, a ' + x) = 0;%restN-zero: ⊦ restn(0, x) = x;%fromE-no: ⊦ a ≠ b → frome(a ' + x, b) = frome(x, b);%pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;%restN-rec: ⊦ restn(n + 1, x) = restn(n, x.tail);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ (a ' + x).tail = x; (s)% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)