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Proof step 5 from proof ex (previous step)

a × d ∈ ax ⊦ ∃ d, ay, az. ax = ay + (a × d)' + az

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ p ∈ ax ↔ (∃ ax0, ax1. ax = ax0 + (p + ax1))
Proof step 7:
∃ ax0, ax1. ax = ax0 + ((a × d) + ax1) ⊦ ∃ d, ay, az. ax = ay + (a × d)' + az

Rule argument

spec: list inst: assoclista name: In seq: ⊦ p ∈ ax ↔ (∃ ax0, ax1. ax = ax0 + (p + ax1)) subst: [ax, p] → [ax, a × d] flags: rewrite

Additional information

Used terms: ax, a × d