Theorem cancel 892

Description: Cancellation law eliminating →₁ consequent. (Contributed by NM, 21-Feb-2002.)

Hypothesis

Ref	Expression
cancel.1	((d ∪ (a →1 c)) →1 c) = ((d ∪ (b →1 c)) →1 c)

Assertion

Ref	Expression
cancel	(d ∪ (a →1 c)) = (d ∪ (b →1 c))

Proof of Theorem cancel

Step	Hyp	Ref	Expression
1		cancel.1	. . 3 ((d ∪ (a →1 c)) →1 c) = ((d ∪ (b →1 c)) →1 c)
2	1	cancellem 891	. 2 (d ∪ (a →1 c)) ≤ (d ∪ (b →1 c))
3	1	ax-r1 35	. . 3 ((d ∪ (b →1 c)) →1 c) = ((d ∪ (a →1 c)) →1 c)
4	3	cancellem 891	. 2 (d ∪ (b →1 c)) ≤ (d ∪ (a →1 c))
5	2, 4	lebi 145	1 (d ∪ (a →1 c)) = (d ∪ (b →1 c))

Syntax hints:   = wb 1   ∪ wo 6   →₁ wi1 12

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by: (None)