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Proof step 2 from proof is-encoded-empty (previous step)

is-encoded(nx, mkbuf(0)) ⊦ nx = []

User applied rule structural induction

And closed the goal!

Rule argument

indvar: nx

Additional information

Used simplifier rules: is-encoded-rec: ⊦ is-encoded(n ' + nx, buf) ↔ (∃ buf0, buf1. buf = buf0 + buf1 ∧ is-encoded(rnode(n), buf0) ∧ is-encoded(nx, buf1));%nat-list freely generated by [], +;;% ⊦ n + nx = n ' + nx; (s)% ⊦ n ' + nx ≠ []; (s)%a: ⊦ (nx + nx0) + nx1 = nx + nx0 + nx1; from specification list[nat-list-set]% ⊦ buf + buf0 = mkbuf(0) ↔ buf = mkbuf(0) ∧ buf0 = mkbuf(0); (ws)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]%is-encoded-size: ⊦ is-encoded(rnd, buf) → # buf = EB_PAGE_SIZE; (forward) from specification encoding-page-size-nonempty[encoding-ref-node]% ⊦ # mkbuf(n) = n; (s)% ⊦ EB_PAGE_SIZE ≠ 0; (s)