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Proof step 4 from proof pos-equal (previous step)

pos(b, a0 ' + x0) = pos(a, x0) + 1, a ∈ x0, ∀ b, a. pos(b, x0) = pos(a, x0) ∧ a ∈ x0 → a = b ⊦ a = b, a = a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1
Proof step 6:
b ≠ a0, pos(b, x0) + 1 = pos(a, x0) + 1, a ∈ x0, ∀ b, a. pos(b, x0) = pos(a, x0) ∧ a ∈ x0 → a = b ⊦ a = b, a = a0

Rule argument

spec: inst: name: pos-n seq: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1 subst: [x, b, a] → [x0, a0, b] flags: rewrite, current

Additional information

Used simplifier rules: pos-y: ⊦ pos(a, a ' + x) = 0;%a: ⊦ (x + y) + z = x + y + z; from specification list%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: x0, a0, b