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Proof step 6 from proof unique-del (previous step)

unique_al(((a × d)' + ax0) -1l (a0 × d0)), ∀ a, d. unique_al(ax0 -1l (a × d)) → a ∈ ax0 -1l (a × d) ∨ unique_al(ax0) ⊦ unique_al((a × d)' + ax0), a0 ∈ ((a × d)' + ax0) -1l (a0 × d0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ p ≠ p0 → (p0 ' + ax) -1l p = p0 ' + (ax -1l p)
Proof step 8:
a0 × d0 ≠ a × d, unique_al((a × d)' + (ax0 -1l (a0 × d0))), ∀ a, d. unique_al(ax0 -1l (a × d)) → a ∈ ax0 -1l (a × d) ∨ unique_al(ax0) ⊦ unique_al((a × d)' + ax0), a0 ∈ (a × d)' + (ax0 -1l (a0 × d0))

Rule argument

seq: ⊦ p ≠ p0 → (p0 ' + ax) -1l p = p0 ' + (ax -1l p) subst: [p, ax, p0] → [a0 × d0, ax0, a × d] rotate: F spec: list-del inst: assoclista name: del1-n

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;% ⊦ (p ' + ax) -1l p = ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; Used terms: a0 × d0, ax0, a × d