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Proof step 2 from proof abs-padding-eq (previous step)

buf0 + buf1 = buf2 + buf3, abs-padding(buf0) ⊦ abs-padding(buf2) ↔ buf0 = buf2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# buf = NODE_HEADER_SIZE, # buf5 = NODE_HEADER_SIZE, (buf5 + buf6 + validtrailer) + buf1 = (buf + buf4 + validtrailer) + buf3, is-aligned(# buf6, 2 * NODE_HEADER_SIZE), is-aligned(# buf4, 2 * NODE_HEADER_SIZE), is-encoded(nodeheader(# buf6, true), buf5), is-encoded(nodeheader(# buf4, true), buf) ⊦ buf5 + buf6 = buf + buf4

Rule argument

Additional information

Used simplifier rules: abs-padding: ⊦ abs-padding(buf) ↔ (∃ buf0, buf1. buf = buf0 + buf1 + validtrailer ∧ is-encoded(nodeheader(# buf1, true), buf0) ∧ is-aligned(# buf1, 2 * NODE_HEADER_SIZE));%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ buf + buf0 = buf1 + buf0 ↔ buf = buf1; (s)%is-encoded-size: ⊦ is-encoded(ndhd, buf) → # buf = NODE_HEADER_SIZE; (forward) from specification encoding-fixed-size-nonempty[encoding-node-header]